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#### Page No 3.106:

Let the length and breadth of the rectangle be and units respectively

Then, area of rectangle square units

If length is increased and breadth reduced each by units, then the area is reduced by square units

Therefore,

Then the length is reduced by unit and breadth is increased by units then the area is increased by square units

Therefore,

Thus we get the following system of linear equation

By using cross multiplication, we have

and

The length of rectangle is units.

The breadth of rectangle is units.

square units

Hence, the area of rectangle is square units

#### Page No 3.106:

Let the length and breadth of the rectangle be and units respectively

Then, area of rectangle =square units

If length is increased by meters and breadth is decreased by meters when the area of a rectangle remains the same

Therefore,

If the length is decreased by meters and breadth is increased by meters when the area remains unaffected, then

Thus we get the following system of linear equation

By using cross-multiplication, we have

and

Hence, the length of rectangle ismeters

The breadth of rectangle is meters

#### Page No 3.106:

Let the length and breadth of the rectangle be and units respectively

Then, area of rectangle = square units

If the length is increased by meters and breath is reduced each by square meters the area is reduced by square units

Therefore,

Then the length is reduced by meter and breadth is increased by meter then the area is increased by square units

Therefore,

Thus, we get the following system of linear equation

By using cross multiplication we have

and

Hence, the length of rectangle ismeter,

The breath of rectangle is meter.

#### Page No 3.106:

Let the income of be Rs and the income of be Rs.further let the expenditure of be and the expenditure of be respectively then,

Saving of =

Saving of =

Solving equation and by cross- multiplication, we have

The monthly income of =

The monthly income of

Hence the monthly income of is Rs

The monthly income of is Rs

#### Page No 3.106:

Let the money with A be Rs x and the money with B be Rs y.

If A gives Rs 30 to B, Then B will have twice the money left with A, According to the condition we have,

If B gives Rs 10 to A, then A will have thrice as much as is left with B,

By multiplying equation with 2 we get,

By subtracting from we get,

By substituting in equation we get

Hence the money with A be and the money with B be

#### Page No 3.106:

Let us take the A examination room will be x and the B examination room will be y

If 10 candidates are sent from A to B, the number of students in each room is same. According to the above condition equation will be

If 20 candidates are sent from B to A, the number of students in A is double the number of students in B, then equation will be,

By subtracting the equationfromwe get,

Substituting in equation, we get

Hence candidates are in A examination Room,

candidates are in B examination Room.

#### Page No 3.107:

A man can alone finish the work in days and one boy alone can finish it in days then

One mans one days work =

One boys one days work=

2men one day work=

7boys one day work=

Since 2 men and 7 boys can finish the work in 4 days

Again 4 men and 4 boys can finish the work in 3 days

Putting and in equation and we get

By using cross multiplication we have

Now,

and

Hence, one man alone can finish the work in and one boy alone can finish the work in .

#### Page No 3.107:

Let , and

Substitute in above equation we get ,

${y}^{°}={7}^{°}+3{x}^{°}$

#### Page No 3.107:

We know that the sum of the opposite angles of cyclic quadrilateral is .in the cyclic quadrilateral, angles and and angles and pairs of opposite angles

Therefore and

Taking

By substituting and we get

Taking

By substituting and we get

By multiplying equation by we get

By subtracting equation from we get

By substituting in equation we get

The angles of a cyclic quadrilateral are

Hence, the angles of cyclic quadrilateral ABCD are .

#### Page No 3.107:

Let take first class full of fare is Rs and reservation charge is Rs per ticket

Then half of the ticket as on full ticket =

According to the given condition we have

Multiplying equation by 2 we have

Subtracting from we get

Putting in equation we get

Hence, the basic first class full fare is

The reservation charge is .

#### Page No 3.107:

We have to prove that the triangle is right

Given and

Sum of three angles in triangle are

By solving with we get,

Multiplying equation by 3 we get

Subtracting equation from

Substituting in equation we get

Angles and are

A right angled triangle is a triangle in which one side should has a right angle that is in it.

Hence, The triangle ABC is right angled

#### Page No 3.107:

Let the fixed charges of car be per km and the running charges be km/hr

According to the given condition we have

Putting in equation we get

Therefore, Total charges for travelling distance of km

= Rs

Hence, A person have to pay for travelling a distance of km.

#### Page No 3.107:

Let the fixed charges of hostel be and the cost of food charges be per day

According to the given condition we have,

Subtracting equation from equation we get

Putting in equation we get

Hence, the fixed charges of hostel is .

The cost of food per day is .

#### Page No 3.107:

Let perimeter of rectangular garden will be .if half the perimeter of a garden will be

When the length is four more than its width then

Substituting in equation we get

Putting in equation we get

Hence, the dimensions of rectangular garden are and

#### Page No 3.107:

We know that the sum of supplementary angles will be.

Let the longer supplementary angles will be.

Then,

If larger of supplementary angles exceeds the smaller by degree, According to the given condition. We have,

Substitute in equation, we get,

Put equation, we get,

Hence, the larger supplementary angle is ,

The smaller supplementary angle is.

#### Page No 3.107:

1 women alone can finish the work in days and 1 man alone can finish it in days .then

One woman one day work=

One man one days work =

2 women’s one days work=

5 man’s one days work =

Since 2 women and 5 men can finish the work in 4 days

3 women and 6 men can finish the work in 3 days

Putting and in equation and we get

By using cross multiplication we have

Now ,

Hence, the time taken by 1 woman alone to finish the embroidery is,

The time taken by 1 man alone to finish the embroidery is .

#### Page No 3.107:

Let the strike money of first cock-owner be and of second cock-owner be respectively. Then we have,

For second cock-owner according to given condition we have,

By subtracting from, we have,

Putting in equation we get,

Hence the stake of money first cock-owner is and of second cock-owner is respectively.

#### Page No 3.107:

Let be the notes of and notes will be

If Meena ask for and notes only, then the equation will be,

Divide both sides by then we get,

If Meena got 25 notes in all then the equation will be,

By subtracting the equation from we get,

Substituting in equation, we get

Therefore and

Hence, Meena has notes of and notes of

#### Page No 3.107:

Hence total number of questions will be

If yash scored marks in atleast getting marks for each right answer and losing mark for each wrong answer then

If 4 marks awarded for each right answer and 2 marks deduced for each wrong answer the he scored 50 marks

By multiplying equation by 2 we get

By subtracting from we get

Putting in equation we have

Total number question will be

Hence, the total number of question is .

#### Page No 3.107:

Let the number of students be and the number of row be .then,

Number of students in each row

Where three students is extra in each row, there are one row less that is when each row has students the number of rows is

Total number of students =no. of rowsno. of students in each row

If three students are less in each row then there are rows more that is when each row has

Therefore, total number of students=Number of rowsNumber of students in each row

Putting in and equation we get

Putting in equation we get

Hence, the number of students in the class is.

#### Page No 3.107:

21. Let the money with first person be and the money with the second person be. Then,

If first person gives to second person then the second person will become twice as rich as first person, According to the given condition, we have,

if second person gives to first person then the first person will becomes six times as rich as second person, According to given condition, we have,

Multiplying equation by we get,

By subtracting from, we get

Putting in equation, we get,

Hence, first person’s capital will be ,

Second person’s capital will be .

#### Page No 3.108:

The given system of equations is

.

For the equations to have no solutions

By cross multiplication we get,

Hence, the value of k is when system equations has no solution.

#### Page No 3.108:

The given systems of equations are

For the equations to have infinite number of solutions,

By cross Multiplication we get,

Hence the value of k is when equations has infinitely many solutions.

#### Page No 3.108:

We know that the sum of the opposite angles of cyclic quadrilateral is .in the cyclic quadrilateral angles and and angles and pairs of opposite angles

Therefore

and

By substituting and we get

Divide both sides of equation by 4 we get

By substituting and we get

By multiplying equation by 3 we get

By subtracting equation from we get

By substituting in equation we get

The angles of a cyclic quadrilateral are

Hence the angles of quadrilateral are

#### Page No 3.109:

The given system of equations are

For the equations to have no solutions,

If we take

Therefore the value of k is10.

Hence, correct choice is.

#### Page No 3.109:

The given system of equations are,

Here,

By cross multiply we get

Therefore the value of k is 6,

Hence, the correct choice is.

#### Page No 3.109:

The given systems of equations are

For the equations to have infinite number of solutions,

Here ,

Let us take

By cross multiplication we get,

Now take

By cross multiplication we get,

Substitute in the above equation

Substitute in equation we get,

Therefore and.

Hence, the correct choice is.

#### Page No 3.109:

The given equations are

For the equations to have infinite number of solutions

Let us take

Hence, the value of when the pair of linear equations has infinitely many solutions.

#### Page No 3.109:

The given equations are

Every solution of the second equation is also a solution of the first equation.

Hence, there are, the system equation is consistent.

#### Page No 3.109:

The given linear pair of equations are

If then

Hence, the number of solutions of the pair of linear equation is.

Therefore, the equations have no solution.

#### Page No 3.109:

The given system of equations are

for unique solution

Here

By cross multiply we get

Hence, the correct choice is.

#### Page No 3.109:

The given system of equations are

For the equations to have infinite number of solutions,

Here,

Therefore

By cross multiplication of we get,

And

Therefore the value of k is 6

Hence, the correct choice is .

#### Page No 3.110:

The given system of equations is inconsistent,

If the system of equations is in consistent, we have

Therefore, the value of k is2.

Hence, the correct choice is .

#### Page No 3.110:

Given the system of equations has

We know that intersecting lines have unique solution

Here

Therefore intersecting lines, have unique solution

Hence, the correct choice is

#### Page No 3.110:

Given the system of equations are

For the equations to have infinite number of solutions,

By cross multiplication we have

Divide both sides by 2. we get

Hence, the correct choice is .

#### Page No 3.110:

The given system of equation is

If then the equation have no solution.

By cross multiply we get

Hence, the correct choice is.

#### Page No 3.110:

The given system of equations are

For coincident lines , infinite number of solution

Option A.:

Option B:

Option.C:

a - b = 0

-5 - (-1) = -4 $\ne$0

None of the option satisfies the values.

#### Page No 3.110:

If a pair of linear equations in two variables is consistent, then its solution exists.

∴The lines represented by the equations are either intersecting or coincident.

Hence, correct choice is.

#### Page No 3.110:

Given the area of the triangle formed by the line

If in the equation either A and B approaches infinity, The line become parallel to either x axis or y axis respectively,

Therefore

Area of triangle

Hence, the correct choice is .

#### Page No 3.110:

Given and

We have plotting points as when

Therefore, area of

Area of triangle is square units

Hence, the correct choice is .

#### Page No 3.110:

The given system of equations

For the equations to have infinite number of solutions

If we take

And

Therefore, the value of k is 6.

Hence, the correct choice is .

#### Page No 3.111:

The given systems of equations are

If

Here

Hence, the correct choice is .

#### Page No 3.111:

Given and

We have plotting points as when

Therefore, area of

Area of triangle is square units

Hence, the correct choice is

#### Page No 3.111:

Given and

If We have plotting points as

Therefore, area of

Area of triangle is square units

Hence, the correct choice is

#### Page No 3.12:

Let no. of ride is   and no. of Hoopla is .He paid Rs 20 for ride and for Hoopla.

The cost of ride is Rs and cost of Hoopla is Rs.then

The number of Hoopla is the half number of ride, then

Hence algebraic equations are and

Now, we draw the graph for algebraic equations.

#### Page No 3.12:

Let age of Aftab is  years and age of his daughter is years. Years ago his age wastimes older as her daughter was. Then

Three years from now, he will be three times older as his daughter will be, then

Hence the algebraic representation are and

#### Page No 3.12:

The given equation are   and.

In order to represent the above pair of linear equation graphically, we need

Two points on the line representing each equation. That is, we find two solutions

of each equation as given below:

We have,

Putting we get

Putting we get

Thus, two solution of equation are

We have

Putting we get

Putting we get

Thus, two solution of equation are

 8 6

Now we plot the pointand and draw a line passing through

These two points to get the graph o the line represented by equation

We also plot the points and and draw a line passing through

These two points to get the graph O the line represented by equation

We observe that the line parallel and they do not intersect anywhere.

#### Page No 3.12:

Gloria is walking the path joining and

Suresh is walking the path joining and

 0 4 5 0

The graphical representations are

#### Page No 3.12:

(i) Given equation are: 5x + 4y + 8 = 0

7x + 6y − 9 = 0

We have And

Thus the pair of linear equation is intersecting.

(ii) Given equation are: 9x + 3y + 12 = 0

18x + 6y + 24 = 0

We have

Thus the pair of linear is coincident lines.

(iii) Given equation are:

We have

Thus the pair of line is parallel lines.

#### Page No 3.12:

(i) Given the linear equation are:

We know that intersecting condition:

Where

Hence the equation of other line is

(ii) We know that parallel line condition is:

Where

Hence the equation is

(iii) We know that coincident line condition is:

Where

Hence the equation is

#### Page No 3.12:

Let the cost of 1 kg of apples be Rs x.

And, cost of 1 kg of grapes = Rs y

According to the question, the algebraic representation is

For ,

The solution table is

 x 50 60 70 y 60 40 20

For 4x + 2y = 300,

The solution table is

 x 70 80 75 y 10 –10 0

The graphical representation is as follows.

#### Page No 3.29:

The given equations are:

Putting in equation we get:

Putting in equation we get:

Use the following table to draw the graph.

 0 3 3 0

Draw the graph by plotting the two points from table.

Graph of the equation:

Putting in equation we get:

Putting in equation we get:

Use the following table to draw the graph.

 0 6 12/5 0

Draw the graph by plotting the two points from the table.

The two lines intersect at point P.

Hence, and is the solution.

#### Page No 3.29:

The given equations are

Putting in equation we get:

Putting in equation we get:

Use the following table to draw the graph.

 0 5 0

Draw the graph by plotting the two points from table.

Graph the equation

Putting in equation we get:

Putting in equation we get:

Use the following table to draw the graph.

 0 5 10/3 0

Draw the graph by plotting the two points from table.

The two lines intersects at point B

Hence is the solution

#### Page No 3.29:

The given equations are

Putting in equation we get:

Putting in equation we get:

Use the following table to draw the graph.

 0 –1/3 –1 0

Draw the graph by plotting the two points from table.

Graph of the equation….

Putting in equation we ge

Putting in equationwe get

Use the following table to draw the graph.

 0 –4 8/3 0

Draw the graph by plotting the two points from table.

The two lines intersect at points P.

Hence is the solution.

#### Page No 3.29:

The given equations are

Putting in equation we get:

Putting in equation we get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Graph of the equation….

Putting in equation we get:

Putting in equationwe get

Use the following table to draw the graph.

 0 7/2 –7/3 0

Draw the graph by plotting the two points from table.

The two lines intersect at points P.

Hence is the solution.

#### Page No 3.29:

The given equations are

Putting in equation we get:

Putting in equation we get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Graph of the equation….

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

The two lines intersect at points P.

Hence is the solution.

#### Page No 3.29:

The given equations are:

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Plotting the two points equation (i) can be drawn.

Graph of the equation….

Putting in equation, we get:

Putting x=2 in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

We see that the two lines are parallel, so they won’t intersect

Hence there is no solution

#### Page No 3.29:

The given equations are

Putting in equation we get:

Putting in equation we get

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Graph of the equation….

Putting in equation we get

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

The two lines intersect at points P.

Hence is the solution

#### Page No 3.29:

The given equations are:

Putting in equation we get:

Putting in equation we get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table

Graph of the equation….

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table

The two lines intersect at points P.

Hence, and is the solution.

#### Page No 3.29:

The given equations are:

Putting in equation we get:

Putting in equationwe get

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Graph of the equation….

Putting in equationwe get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

The two lines intersect at points P.

Hence, and is the solution.

#### Page No 3.29:

The given equations are

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Graph of the equation….

Putting in equationwe get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

The two lines intersects at points P

Hence, and is the solution.

#### Page No 3.29:

The given equations are

Putting in equation we get:

Putting in equation we get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Graph of the equation….

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Thus the graph of the two equations coincide

Consequently, every solution of one equation is a solution of the other.

Hence the equations have infinitely many solutions.

#### Page No 3.29:

The given equations are

Putting in equation we get:

Putting in equationswe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table

Graph of the equation….

Putting in equations we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Thus the graph of the two equations coincide

Consequently, every solution of one equation is a solution of the other.

Hence the equations have infinitely many solutions.

#### Page No 3.29:

The given equations are

Putting in equation we get:

Putting in equationswe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Graph of the equation….

Putting in equations we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Thus the graph of the two equations coincide

Consequently, every solution of one equation is a solution of the other.

Hence the equations have infinitely many solutions.

#### Page No 3.29:

The given equations are

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Graph of the equation….

Putting in equation we get:

Putting in equationwe get

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Thus the graph of the two equations are coincide

Consequently, every solution of one equation is a solution of the other.

Hence the equations have infinitely many solutions.

#### Page No 3.29:

The given equations are

Putting in equation we get:

Putting in equationwe get

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Graph of the equation….

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Here we see that the two lines are parallel

Hence the given system of equations has no solution.

#### Page No 3.29:

The given equations are

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

The graph of (i) can be obtained by plotting the two points .

Graph of the equation….

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Here the two lines are parallel and so there is no point in common

Hence the given system of equations has no solution.

#### Page No 3.29:

The given equations are

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Here two lines are parallel and so don’t have common points

Hence the given system of equations has no solution.

#### Page No 3.29:

The given equations are

Putting in equation we get:

Putting in equation we get:

Use the following table to draw the graph.

The graph of (i) can be obtained by plotting the two points.

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Here, the two lines are parallel.

Hence the given system of equations is inconsistent.

#### Page No 3.29:

(i) Draw the 3 lines as given by equations

By taking x=1 = 1 cm on x−axis

And y =1=1cm on y−axis

Clearly from graph points of intersection three lines are

(−4,2) , (1,3), (2,5)

(ii) Draw the 3 lines as given by equations

By taking x=1 = 1 cm on x−axis

And y =1=1cm on y−axis

From graph point of intersection are (0,0) (10/3,0) (5/3,5/3)

#### Page No 3.29:

The given equations are

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

 0 5/4 –5/2 0

Draw the graph by plotting the two points from table.

It has unique solution.

Hence the system of equations is consistent

#### Page No 3.29:

(i) The given equations are

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Graph of the equation….

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

The two lines intersect at point .

Hence the equations have unique solution.

(ii) The equations of graphs is

Putting in equation we get:

Putting in equation we get:

Use the following table to draw the graph.

The graph of (i) can be obtained by plotting the two points .

Graph of the equation

Putting in equation we get.

Putting in equation we get.

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

The two lines are coincident.

Hence the equations have infinitely much solution.

Hence the system is consistent

#### Page No 3.29:

(i) The given equations are

The two points satisfying (i) can be listed in a table as,

 −2 8 0 4

The two points satisfying (ii) can be listed in a table as,

 4 2 0 4

Now, graph of equations (i) and (ii) can be drawn as,

It is seen that the solution of the given system of equations is given by x = 3, y = 2.

Also, it is observed that the lines (i) and (ii) meet the y-axis at the points respectively.

(ii) The given equations are

The two points satisfying (i) can be listed in a table as,

 4 6 0 –3

The two points satisfying (ii) can be listed in a table as,

 3 2 5.5 3

Now, graph of equations (i) and (ii) can be drawn as,

It is seen that the solution of the given system of equations is given by x = 2, y = 3.

Also, it is observed that the lines (i) and (ii) meet the y-axis at the points respectively.

(iii) The given equations are

The two points satisfying (i) can be listed in a table as,

 3 1 5 9

The two points satisfying (ii) can be listed in a table as,

 1 5 0 4

Now, graph of equations (i) and (ii) can be drawn as,

It is seen that the solution of the given system of equations is given by x = 4, y = 3.

Also, it is observed that the lines (i) and (ii) meet the y-axis at the points respectively.

(iv) The given equations are

The two points satisfying (i) can be listed in a table as,

 5 7 1 0

The two points satisfying (ii) can be listed in a table as,

 2 1 0 –2

Now, graph of equations (i) and (ii) can be drawn as,

It is seen that the solution of the given system of equations is given by x = 3, y = 2.

Also, it is observed that the lines (i) and (ii) meet the y-axis at the points respectively.(v) The given equations are

The two points satisfying (i) can be listed in a table as,

 1 3 2 –4

The two points satisfying (ii) can be listed in a table as,

 1 4 –3 3

Now, graph of equations (i) and (ii) can be drawn as,

It is seen that the solution of the given system of equations is given by x = 2, y = −1.

Also, it is observed that the lines (i) and (ii) meet the y-axis at the points respectively.

(vi) The given equations are

The two points satisfying (i) can be listed in a table as,

 1 3 –3 1

The two points satisfying (ii) can be listed in a table as,

 1 5 –2 2

Now, graph of equations (i) and (ii) can be drawn as,

It is seen that the solution of the given system of equations is given by x = 2, y = −1.

Also, it is observed that the lines (i) and (ii) meet the y-axis at the points respectively.

#### Page No 3.30:

(i) The given equations are

The two points satisfying (i) can be listed in a table as,

 0 1 0 1

The two points satisfying (ii) can be listed in a table as,

 1 3 2 6

The two points satisfying (iii) can be listed in a table as,

 0 6 6 6

Now, graph of equations (i), (ii) and (iii) can be drawn as,

It is seen that the coordinates of the vertices of the obtained triangle are

(ii) The given equations are

The two points satisfying (i) can be listed in a table as,

 0 2 0 2

The two points satisfying (ii) can be listed in a table as,

 3 –3 1 –1

The two points satisfying (iii) can be listed in a table as,

 3 5 5 3

Now, graph of equations (i), (ii) and (iii) can be drawn as,

It is seen that the coordinates of the obtained triangle are

#### Page No 3.30:

The given equations are:

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

The two lines intersect at. The region enclosed by the lines represented by the given equations and x−axis are shown in the above figure

Hence, and is the solution.

(ii) The given equations are:

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

 0 2 0

The graph of (i) can be obtained by plotting the two points.

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

 −

Draw the graph by plotting the two points from table.

The two lines intersect at. The area enclosed by the lines represented by the given equations and the coordinates x−axis and shaded the area in graph.

Hence, and is the solution.

(iii) The given equations are:

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

The two lines intersect at. The area enclosed by the lines represented by the given equations and the coordinates x−axis and shaded the area in graph.

Hence, and is the solution.

#### Page No 3.30:

The given equations are

Putting in equation we get:

Putting in equation we get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Draw the graph by plotting the two points from table.

The intersection point is P(3, 2)

Three points of the triangle are

Hence the value of and

#### Page No 3.30:

The given equations are

Putting in equation we get:

Putting in equation (i) we get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

The two lines intersect at.

Now, Required area = Area of shaded region

Required area = Area of PBD

Required area =

Required area =

Required area =

Hence the area =

#### Page No 3.30:

The given equations are

Putting in equation we get:

Putting in equation we get:

Use the following table to draw the graph.

The graph of (i) can be obtained by plotting the points (0, 4/3), (−1, 0).

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

The two lines intersect at.

Hence is the solution of the given equations.

Now,

Required area = Area of PBD

Required area =

Required area =

Required area =

Hence, the area =

#### Page No 3.30:

The given equations are

Putting in equation we get:

Putting in equation we get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

The two lines intersect at.

Hence is the solution of the given equations

The area enclosed by the lines represented by the given equations and the y−axis is shaded region in the figure

Now, Required area = Area of shaded region

Required area = Area of PAC

Required area =

Required area =

Required area =

Hence the required area is sq. unit

#### Page No 3.30:

(i)

The given equations are

The two points satisfying (i) can be listed in a table as,

 4 0 −2 6

The two points satisfying (ii) can be listed in a table as,

 4 6 3 4

Now, graph of equations (i) and (ii) can be drawn as,

It is seen that the solution of the given system of equations is given by x = 2, y = 2.

Also, it is observed that the coordinates of the points where the lines (i) and (ii) meet the x-axis are respectively.

(ii) The given equations are

The two points satisfying (i) can be listed in a table as,

 0 2 2 2

The two points satisfying (ii) can be listed in a table as,

 1 3 –4 4

Now, graph of equations (i) and (ii) can be drawn as,

It is seen that the solution of the given system of equations is given by x = 3, y = 4.

Also, it is observed that the coordinates of the points where the lines (i) and (ii) meet the x-axis are respectively.

(iii) The given equations are

The two points satisfying (i) can be listed in a table as,

 0 −1 2.5 3

The two points satisfying (ii) can be listed in a table as,

 4 –5 4 –2

Now, graph of equations (i) and (ii) can be drawn as,

It is seen that the solution of the given system of equations is given by x = 1, y = 2.

Also, it is observed that the coordinates of the points where the lines (i) and (ii) meet the x-axis are respectively.

Solution is missing

(iv) The given equations are

The two points satisfying (i) can be listed in a table as,

 –2 7 4 –2

The two points satisfying (ii) can be listed in a table as,

 –1 3 1 3

Now, graph of equations (i) and (ii) can be drawn as,

It is seen that the solution of the given system of equations is given by x = 1, y = 2.

Also, it is observed that the coordinates of the points where the lines (i) and (ii) meet the x-axis are respectively.

#### Page No 3.30:

The given equations are

The two points satisfying (i) can be listed in a table as,

 −3 6 0 6

The two points satisfying (ii) can be listed in a table as,

 0 9 6 0

The two points satisfying (iii) can be listed in a table as,

 −1 8 2 2

Now, graph of equations (i), (ii) and (iii) can be drawn as,

It is seen that the coordinates of the vertices of the obtained triangle are

Area of ΔABC =

#### Page No 3.30:

The given equations are:

Putting in equation (i) we get:

Putting in equation (i) we get:

Use the following table to draw the graph

Draw the graph by plotting the two points from table.

Putting in equationwe get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

The two lines intersect at.

Hence is the solution of the given equations.

The area enclosed by the lines represented by the given equations and the y−axis

Now,

Required area = Area of PCA

Required area =

Required area =

Required area =

Hence the required area is

#### Page No 3.31:

The given equations are:

Putting in equation we get:

Putting in equation (i) we get:

Use the following table to draw the graph.

 x 0 5 y 0

Draw the graph by plotting the two points from table

Putting in equation (ii) we get:

Putting in equation (ii) we get:

Use the following table to draw the graph.

 x 0 5 y 3 0

Draw the graph by plotting the two points from table.

The three vertices of the triangle are

Hence the solution of the equation is and

#### Page No 3.31:

The given equations are:

Putting in equation (i) we get:

Putting in equation (i) we get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

 –3

Draw the graph by plotting the two points from table.

Hence the vertices of the required triangle are.

Now,

Required area = Area of PCA

Required area =

Required area =

Hence the required area is

#### Page No 3.31:

(i) Let the number of girls be x and the number of boys be y.

According to the question, the algebraic representation is

x + y = 10

xy = 4

For x + y = 10,

x = 10 − y

 x 5 4 6 y 5 6 4

For xy = 4,

x = 4 + y

 x 5 4 3 y 1 0 −1

Hence, the graphic representation is as follows.

From the figure, it can be observed that these lines intersect each other at point (7, 3).

Therefore, the number of girls and boys in the class are 7 and 3 respectively.

(ii) Let the cost of 1 pencil be Rs x and the cost of 1 pen be Rs y.

According to the question, the algebraic representation is

5x + 7y = 50

7x + 5y = 46

For 5x + 7y = 50,

 x 3 10 − 4 y 5 0 10

7x + 5y = 46

 x 8 3 − 2 y − 2 5 12

Hence, the graphic representation is as follows.

From the figure, it can be observed that these lines intersect each other at point (3, 5).

Therefore, the cost of a pencil and a pen are Rs 3 and Rs 5 respectively.

(iii) Let us denote the number of pants by x and the number of skirts by y. Then the equations formed are :

y = 2x − 2 … (i)

y = 4x − 4 … (ii)

The graphs of the equations (i) and (ii) can be drawn by finding two solutions for each of the equations. They are given in the following table.

 x 2 0 y = 2x − 2 2 −2

Hence, the graphic representation is as follows.

The two lines intersect at the point (1, 0). So, x = 1, y = 0 is the required solution of the pair of linear equations, i.e., the number of pants she purchased is 1 and she did not buy any skirt.

#### Page No 3.31:

The given equations are:

Putting in equation (i) we get:

Putting in equation (i) we get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Putting in equation (ii) we get:

Putting in equation (ii) we get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

The two lines intersect at points of y−axis.

Hence, and is the Solution.

(ii) The equations are:

Putting in equation (1) we get:

Putting in equation (1) we get:

Use the following table to draw the graph:

Draw the graph by plotting the two points from table.

Putting in equation (2) we get:

Putting in equation (2) we get:

Use the following table to draw the graph.

 x

Draw the graph by plotting the two points from table.

Two lines intersect at points of y−axis.

Hence and is the solution.

#### Page No 3.31:

The given equations are

Putting in equation (i) we get:

Putting in equationwe get:

Use the following table to draw the graph.

 x

The graph of (i) can be obtained by plotting the two points.

Putting in equation (ii) we get:

Putting in equation (ii) we get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Graph of lines represented by the equations meet y−axis at respectively.

#### Page No 3.31:

(i) For intersecting lines,

Equation of another intersecting line to the given line is−

Since, condition for intersecting lines and unique solution is−

(ii) For parallel lines,

Equation of another parallel line to the given line is−

Since, condition for parallel lines and no solution is−

(iii) For co−incident lines,

Equation of another coincident line to the given line is−

Since, condition for coincident lines and infinite solution is−

#### Page No 3.43:

The given equations are:

Multiply equation by 2 and equation by 15, and add both equations we get

Put the value of in equationwe get

Hence the value of and

#### Page No 3.43:

The given equations are:

Multiply equation by 2 and equation by, and add both equations we get

Put the value of in equationwe get

Hence the value of and

#### Page No 3.43:

The given equations are:

Multiply equation by 2 and equation by, and add both equations we get

Put the value of in equationwe get

Hence the value of and

#### Page No 3.43:

The given equations are:

Subtract (ii) from (i) we get

Put the value of in equation we get

Hence the value of and .

#### Page No 3.43:

The given equations are:

Multiply equation by and equation by and add both equations we get

Put the value of in equationwe get

Hence the value of and

#### Page No 3.43:

The given equations are:

Multiply equation by and add both equations we get

Put the value of in equation we get

Hence the value of and .

#### Page No 3.43:

The given equations are:

Multiply equation by and equation by and add both equations we get

Put the value of in equation we get

Hence the value of and .

#### Page No 3.43:

The given equations are:

Multiply equation by and equation by and add both equations we get

Put the value of in equation we get

Hence the value of and .

#### Page No 3.44:

The given equations are:

Multiply equation by and subtract equations, we get

Put the value of in equation, we get

Hence the value of x and y are and

#### Page No 3.44:

The given equations are:

Multiply equation by and subtract equation (ii) from (i) we get

Put the value of in equation, we get

Hence the value of and

#### Page No 3.44:

The given equations are:

Multiply equation by and equation by and subtract equation (ii) from (i), we get

Put the value of in equation, we get

Hence the value of and

#### Page No 3.44:

The given equations are:

Multiply equation (1) by , we get

adding (2) and (3), we get

Hence the value of x and y are and

#### Page No 3.44:

The given equations are:

Multiply equation by and by and subtract equation (ii) from (i) we get

Put the value of in equation, we get

Hence the value of and

#### Page No 3.44:

The given equations are:

Multiply equation by and by and subtract equation (ii) from (i) we get

Put the value of in equation, we get

Hence the value of and

#### Page No 3.44:

The given equations are:

Multiply equation by and add both equations we get

Put the value of in equation, we get

Hence the value of and

#### Page No 3.44:

The given equations are:

Multiply equation by and by and subtract equation (ii) from (i) we get

Put the value of in equation, we get
$\frac{1}{2×\frac{1}{2}}+\frac{1}{3y}=2\phantom{\rule{0ex}{0ex}}\frac{1}{3y}=2-1\phantom{\rule{0ex}{0ex}}\frac{1}{3y}=1\phantom{\rule{0ex}{0ex}}y=\frac{1}{3}\phantom{\rule{0ex}{0ex}}$

Hence the value of and $y=\frac{1}{3}$

#### Page No 3.44:

The given equations are:

Put the value of in equation, we get

Hence the value of and

#### Page No 3.44:

The given equations are:

Multiply equation by and subtract (ii) from (i), we get

Put the value of in equation, we get

Hence the value of and .

#### Page No 3.44:

The given equations are:

Multiply equation by and add both equations, we get

Put the value of in equation, we get

Hence the value of and

#### Page No 3.44:

The given equations are:

Multiply equation by and subtract (ii) from equation (i), we get

Put the value of in equation, we get

Hence the value of and .

#### Page No 3.44:

The given equations are:

Multiply equation by and equation by, add both equations, we get

Put the value of in equation, we get

Hence the value of and

#### Page No 3.44:

The given equations are:

Multiply equation by and subtract (ii) from (i), we get

Put the value of in equation, we get

Hence the value of and

#### Page No 3.44:

The given equations are:

Let and then equations are

Multiply equation by and subtract (ii) from (i), we get

Put the value of in equation, we get

Then

Put the value of in second equation, we get

Hence the value of and.

#### Page No 3.44:

The given equations are:

Put the value of in equation, we get

Hence the value of and

#### Page No 3.44:

The given equations are:

Let and then equations are

Multiply equation by and subtracting (ii) from (i), we get

Put the value of in equation, we get

Then

Put the value of in second equation, we get

Hence the value of and .

#### Page No 3.44:

The given equations are:

Let and then equations are

Multiply equation by and equation by and add both equations, we get

Put the value of in equation, we get

Then

Put the value of in first equation, we get

Hence the value of and .

#### Page No 3.45:

The given equations are:

Let and then equations are

Multiply equation by and add both equations, we get

Put the value of in equation, we get

Then

Put the value of in first equation, we get

Hence the value of and .

#### Page No 3.45:

The given equations are:

Let and then equations are

Multiply equation by and equation by add both equations, we get

Put the value of in equation, we get

Then

Put the value of in equation we get

Hence the value of and .

#### Page No 3.45:

The given equations are:

Let and then equations are

Put the value of in equation, we get

Then

Hence the value of and .

#### Page No 3.45:

The given equations are:

Multiply equation by and subtract (ii) from (i), we get

Put the value of in equation, we get

Hence the value of and

#### Page No 3.45:

The given equations are:

Put the value of in equation, we get

Hence the value of and .

#### Page No 3.45:

The given equations are:

Multiply equation by and add both equations, we get

Put the value of in equation, we get

Hence the value of and .

#### Page No 3.45:

The given equations are:

Let and then equations are

Multiply equation by and subtract (ii) from (i), we get

Put the value of in equation, we get

Then

Put the value of in equation we get

Hence the value of and .

#### Page No 3.45:

The given equations are:

Multiply equation by and equation by, add both equations, we get

Put the value of in equation, we get

Hence the value of and .

#### Page No 3.45:

The given equations are:

Multiply equation by and equation by, and subtract (ii) from (i) we get

Put the value of in equation, we get

Hence the value of and

#### Page No 3.45:

The given equations are:

Multiply equation by and equation by and subtract (ii) from (i) we get

Put the value of in equation, we get

Hence the value of and

#### Page No 3.45:

The given equations are:

First of all we find the value of

Put the value of in equation, we get

Put the value of and in equation in we get

Multiply equation by and add equations and, we get

Put the value of in equation, we get

Put the value of and in equation we get

Hence the value of, and.

#### Page No 3.45:

The given equations are:

First of all we find the value of

Put the value of in equation, we get

Put the value of and in equation in we get

Put the value of and in equation, we get

Hence the value of, and.

#### Page No 3.45:

The given equations are:

Let and then equations are

Multiply equation by and equation by add both equations, we get

Put the value of in equation, we get

Then

Put the value of in equation we get

Hence the value of and

#### Page No 3.45:

The given equations are:

Multiply equation by and equation by and subtract (ii) from (i) we get

Put the value of in equation, we get

Hence the value of and

#### Page No 3.45:

The given equations are:

Multiply equation by and equation by 3 and add both equations we get

Put the value of in equation, we get

Hence the value of and .

#### Page No 3.45:

The given equations are:

Let and then equations are

Multiply equation by and add both equations, we get

Put the value of in equation, we get

Then

Hence the value of and .

#### Page No 3.45:

The given equations are:

Let and then equations are

Multiply equation by and equation by and add both equations, we get

Put the value of in equation, we get

Then

Put the value of in equation we get

Hence the value of and .

#### Page No 3.45:

The given equations are:

Let and then equations are

Multiply equation by and add both equations, we get

Put the value of in equation, we get

Then

Put the value of in equation we get

Hence the value of and

#### Page No 3.45:

The given equations are:

Multiply equation by and add both equations we get

Put the value of in equation, we get

Hence the value of and

#### Page No 3.45:

The given equations are:

Multiply equation by and equation by, add both equations we get

Put the value of in equation, we get

Hence the value of and

#### Page No 3.45:

The given equations are:

Multiply equation by and equation by and add both equations we get

Put the value of in equation, we get

Hence the value of and

#### Page No 3.56:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

By cross multiplication method we get

and

Hence we get the value of and

#### Page No 3.56:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

By cross multiplication method we get

Also

Hence we get the value of and

#### Page No 3.56:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

Also

Hence we get the value of and

#### Page No 3.56:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

Hence we get the value of and

#### Page No 3.56:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

Let

By cross multiplication method we get

So

We know that

Hence we get the value of and

#### Page No 3.56:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

T

Thereforeand

Hence we get the value of and

#### Page No 3.56:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

, and

Hence we get the value of and

#### Page No 3.56:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

And

Hence we get the value of and

#### Page No 3.56:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

So for x we have

And

Hence we get the value of and

#### Page No 3.56:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

And

Hence we get the value of and

#### Page No 3.56:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

And

Hence we get the value ofand

#### Page No 3.56:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

Rewriting the equation again

By cross multiplication method we get

And

We know that

Substituting value of x in equation (3) we get

Hence we get the value ofand

#### Page No 3.56:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

Rewriting the equation again

By cross multiplication method we get from eq. (1) and eq. (2)

And

We know that

Hence we get the value of and

#### Page No 3.56:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

And

Hence we get the value of and

#### Page No 3.56:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

Now consider

And

Hence we get the value of and

#### Page No 3.56:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

Consider the following to calculate x

And

Hence we get the value of and

#### Page No 3.56:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

And

Hence we get the value of and

#### Page No 3.56:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

Consider the following for x

And

Hence we get the value of and

#### Page No 3.56:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

Consider the following for x

Now for y

Hence we get the value of and

#### Page No 3.56:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

Consider the following for x

Now consider the following for y

Hence we get the value of and

#### Page No 3.56:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

Consider the following for x

Now consider the following for y

Hence we get the value of and

#### Page No 3.56:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

Now rewriting the given equation as

By cross multiplication method we get

Consider the following for u

Consider the following for v

We know that

Now adding eq. (3) and (4) we get

And after substituting the value of x in eq. (4) we get

Hence we get the value of and

#### Page No 3.56:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

After rewriting equations

By cross multiplication method we get

For y consider the following

Hence we get the value of and

#### Page No 3.57:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation, after rewriting equations

By cross multiplication method we get

Consider the following for y

Hence we get the value of and

#### Page No 3.57:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

Let

Rewriting equations

Now, by cross multiplication method we get

For u consider the following

For y consider

We know that

Now

Hence we get the value of and

#### Page No 3.57:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

Now for y

Hence we get the value of and

#### Page No 3.57:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

For y

Hence we get the value of and

#### Page No 3.57:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

Hence we get the value of

#### Page No 3.71:

GIVEN:

To find: To determine whether the system has a unique solution, no solution or infinitely many solutions

We know that the system of equations

For unique solution

For no solution

For infinitely many solution

Here,

Since which means hence the system of equation has no solution.

Hence the system of equation has no solution

#### Page No 3.71:

GIVEN:

To find: To determine whether the system has a unique solution, no solution or infinitely many solutions

We know that the system of equations

For unique solution

For no solution

For infinitely many solution

Here,

Since which means hence the system of equation has infinitely many solution.

Hence the system of equation has infinitely many solutions

#### Page No 3.71:

GIVEN:

To find: To determine whether the system has a unique solution, no solution or infinitely many solutions

We know that the system of equations

For unique solution

For no solution

For infinitely many solution

Here,

Since which means hence the system of equation has infinitely many solution.

Hence the system of equation has infinitely many solutions

#### Page No 3.71:

GIVEN:

To find: To determine whether the system has a unique solution, no solution or infinitely many solutions

We know that the system of equations

For unique solution

For no solution

For infinitely many solution

Here,

Since which means hence the system of equation has no solution.

Hence the system of equation has no solution

#### Page No 3.71:

GIVEN:

To find: To determine to value of k for which the system has a unique solution.

We know that the system of equations

For unique solution

Here,

Hence for the system of equation has unique solution.

#### Page No 3.71:

GIVEN:

To find: To determine to value of k for which the system has a unique solution.

We know that the system of equations

For unique solution

Here,

Hence for the system of equation has unique solution

#### Page No 3.71:

GIVEN:

To find: To determine to value of k for which the system has a unique solution.

We know that the system of equations

For unique solution

Here,

Hence already for the system of equation to have unique solution but the value of k should be a real number

Hence for the system of equation has unique solution.

#### Page No 3.71:

GIVEN:

To find: To determine to value of k for which the system has a unique solution.

We know that the system of equations

For unique solution

Here,

Hence for the system of equation has unique solution

#### Page No 3.72:

GIVEN:

To find: To determine for what value of k the system of equation has infinitely many solutions

We know that the system of equations

For infinitely many solution

Here,

Hence for the system of equation have infinitely many solutions

#### Page No 3.72:

GIVEN:

To find: To determine for what value of k the system of equation has infinitely many solutions

We know that the system of equations

For infinitely many solution

Here,

Hence for the system of equation have infinitely many solutions.

#### Page No 3.72:

GIVEN:

To find: To determine for what value of k the system of equation has infinitely many solutions

We know that the system of equations

For infinitely many solution

Here,

Hence for the system of equation have infinitely many solutions.

#### Page No 3.72:

GIVEN:

To find: To determine for what value of k the system of equation has infinitely many solutions

We know that the system of equations

For infinitely many solution

Here,

Hence for the system of equation have infinitely many solutions

#### Page No 3.72:

GIVEN:

To find: To determine for what value of k the system of equation has infinitely many solutions

We know that the system of equations

For infinitely many solution

Here

Consider the following

Now consider the following

Hence for the system of equation have infinitely many solutions.

#### Page No 3.72:

GIVEN:

To find: To determine for what value of k the system of equation has infinitely many solutions

We know that the system of equations

For infinitely many solution

Here,

Consider the following for k

Now consider the following

Hence for the system of equation have infinitely many solutions.

#### Page No 3.72:

GIVEN:

To find: To determine for what value of k the system of equation has infinitely many solutions

We know that the system of equations

For infinitely many solution

Here,

Hence for the system of equation have infinitely many solutions.

#### Page No 3.72:

GIVEN:

To find: To determine for what value of k the system of equation has infinitely many solutions

We know that the system of equations

For infinitely many solution

Here,

Consider the following relation to find k

Now consider the following

Hence for the system of equation have infinitely many solutions

#### Page No 3.72:

GIVEN:

To find: To determine for what value of k the system of equation has infinitely many solutions

We know that the system of equations

For infinitely many solution

Here,

Consider the following relation to find k

Now again consider the following

Hence for the system of equation have infinitely many solutions

#### Page No 3.72:

GIVEN:

To find: To determine for what value of k the system of equation has infinitely many solutions

We know that the system of equations

For infinitely many solution

Here

Now again consider the following to find k

Hence for the system of equation have infinitely many solutions

#### Page No 3.72:

GIVEN:

To find: To determine for what value of k the system of equation has infinitely many solutions

We know that the system of equations

For infinitely many solution

Here,

Consider the following to find out k

Now again consider the following relation

So the common solution is 7

Hence for the system of equation have infinitely many solutions

#### Page No 3.72:

GIVEN:

To find: To determine for what value of k the system of equation has no solution

We know that the system of equations

For no solution

Here,

Hence for the system of equation have infinitely many solutions.

#### Page No 3.72:

GIVEN:

To find: To determine for what value of k the system of equation has no solution

We know that the system of equations

For no solution

Here,

Hence for the system of equation has no solution

#### Page No 3.72:

GIVEN:

To find: To determine for what value of k the system of equation has no solution

We know that the system of equations

For no solution

Here,

Hence for the system of equation has no solution.

#### Page No 3.72:

GIVEN:

To find: To determine for what value of k the system of equation has no solution

We know that the system of equations

For no solution

Here,

Hence for the system of equation has no solution.

#### Page No 3.72:

GIVEN:

To find: To determine for what value of k the system of equation has no solution

We know that the system of equations

For no solution

Here,

Hence for the system of equation has no solution.

#### Page No 3.72:

GIVEN:

To find: To determine for what value of c the system of equation has no solution

We know that the system of equations

For no solution

Here,

Hence for the system of equation has no solution.

#### Page No 3.72:

GIVEN:

To find: To determine for what value of k the system of equation will be inconsistent

We know that the system of equations

For the system of equation to be inconsistent

Here,

Hence for the system of equation will be inconsistent.

#### Page No 3.72:

GIVEN:
$\alpha x+3y=\alpha -3\phantom{\rule{0ex}{0ex}}12x+\alpha y=\alpha$
To find: To determine for what value of k the system of equation has no solution

We know that the system of equations

For no solution

Here,

$\frac{\alpha }{12}=\frac{3}{\alpha }\ne \frac{\alpha -3}{\alpha }$
Consider the following for α

$\frac{\alpha }{12}=\frac{3}{\alpha }\phantom{\rule{0ex}{0ex}}{\alpha }^{2}=12×3\phantom{\rule{0ex}{0ex}}{\alpha }^{2}=36\phantom{\rule{0ex}{0ex}}\alpha =±6$

Now consider the following

$\frac{3}{\alpha }\ne \frac{\alpha -3}{\alpha }\phantom{\rule{0ex}{0ex}}3\alpha \ne \alpha \left(\alpha -3\right)\phantom{\rule{0ex}{0ex}}3\alpha \ne {\alpha }^{2}-3\alpha \phantom{\rule{0ex}{0ex}}6\alpha \ne {\alpha }^{2}\phantom{\rule{0ex}{0ex}}\alpha \ne 6$
Hence the common value of α is − 6

Hence for α = -6 the system of equation has no solution

#### Page No 3.72:

GIVEN:

To find: To determine for what value of k the system of equation has

(1) Unique solution

(2) No solution

We know that the system of equations

(1) For Unique solution

Here,

Hence for the system of equation has unique solution.

(2) For no solution

Hence for the system of equation has no solution

#### Page No 3.73:

GIVEN:

To find: To determine for what value of c the system of equation has infinitely many solution

We know that the system of equations

For infinitely many solution

Here

Consider the following

Now consider the following for c

But it is given that c 0. Hence c = 6

Hence for the system of equation have infinitely many solutions.

#### Page No 3.73:

GIVEN:

To find: To determine for what value of k the system of equation has

(1) Unique solution

(2) No solution

(3) Infinitely many solution

We know that the system of equations

(1) For Unique solution

Here,

Hence for the system of equation has unique solution

(2) For no solution

Here,

Hence for the system of equation has no solution

(3) For infinitely many solution

Here,

But since here

Hence the system does not have infinitely many solutions.

#### Page No 3.73:

GIVEN:

To find: To determine for what value of k the system of equation will represents coincident lines

We know that the system of equations

For the system of equation to represent coincident lines we have the following relation

Here,

Hence for the system of equation represents coincident lines

#### Page No 3.73:

GIVEN:

To find: To determine the condition for the system of equation to have a unique equation

We know that the system of equations

For unique solution

Here

Hence for the system of equation have unique solution.

#### Page No 3.73:

GIVEN:

To find: To determine for what value of k the system of equation has infinitely many solutions

We know that the system of equations

For infinitely many solution

Here

Again consider

Hence for and the system of equation has infinitely many solution.

#### Page No 3.73:

GIVEN:

To find: To determine for what value of k the system of equation has infinitely many solutions

We know that the system of equations

For infinitely many solution

Here

Consider the following

5a − 4b + 21 = 0……(1)

Again

a + b + 6 = 0…… (2)

Multiplying eq. (2) by 4 and adding eq. (1)

Putting the value of a in eq. (2)

Hence for and the system of equation has infinitely many solution.

#### Page No 3.73:

GIVEN:

To find: To determine for what value of k the system of equation has infinitely many solutions

We know that the system of equations

For infinitely many solution

Here

…… (1)

Again consider

…… (2)

Multiplying eq. (2) by 3 and subtracting from eq. (1)

Putting the value of q in eq. (2)

Hence for and the system of equation has infinitely many solution.

#### Page No 3.73:

(i) GIVEN:

To find: To determine for what value of k the system of equation has infinitely many solutions

We know that the system of equations

For infinitely many solution

Here

Consider

Again consider

Hence forand the system of equation has infinitely many solution.

(ii) GIVEN:

To find: To determine for what value of k the system of equation has infinitely many solutions

We know that the system of equations

For infinitely many solution

Here

Consider the following

Again consider

Hence for and the system of equation has infinitely many solution.

(iii) GIVEN:

To find: To determine for what value of k the system of equation has infinitely many solutions

We know that the system of equations

For infinitely many solution

Here

Consider the following

Again consider

Hence forand the system of equation has infinitely many solution.

(iv) GIVEN:

To find: To determine for what value of k the system of equation has infinitely many solutions

We know that the system of equations

For infinitely many solution

Here

Consider the following

…… (1)

Again consider

…… (2)

Multiplying eq. (2) by 2 and subtracting eq. (1) from eq. 2

Substituting the value of ‘a’ in eq. (2) we get

Hence forand the system of equation has infinitely many solution.

(v) GIVEN:

To find: To determine for what value of k the system of equation has infinitely many solutions

We know that the system of equations

For infinitely many solution

Here

Consider the following

…… (1)

Again consider the following

…… (2)

Multiplying eq. (2) by 2 and subtracting eq. (1) from eq. (2)

Substituting the value of b in eq. (2) we get

Hence forand the system of equation has infinitely many solution.

(vi) GIVEN:

To find: To determine for what value of k the system of equation has infinitely many solutions

Rewrite the given equations

We know that the system of equations

For infinitely many solution

Here

Consider the following

Hence for the system of equation have infinitely many solutions.

(vii) GIVEN :

To find: To determine for what value of k the system of equation has infinitely many solutions

We know that the system of equations

For infinitely many solution

Here

Consider the following

Hence for the system of equation have infinitely many solutions.
(viii) Given:

We know that the system of equations

has infinitely many solutions if

So,

Solving these two equations, we get
$-4b=-4\phantom{\rule{0ex}{0ex}}⇒b=1$
Putting b = 1 in a + b = 4, we get
a = 3

(ix) Given:
$2x+3y=7\phantom{\rule{0ex}{0ex}}2ax+ay=28-by$
$⇒2ax+\left(a+b\right)y=28$
We know that the system of equations

has infinitely many solutions if

Now,
$\frac{1}{a}=\frac{3}{a+b}$
⇒ a + b = 3a
b = 2a              .....(1)
Also, $\frac{3}{a+b}=\frac{1}{4}$
a + b = 12     .....(2)
Solving (1) and (2), we get
a = 4 and b = 8

#### Page No 3.77:

To find:

(1) Total amount of A.

(2) Total amount of B.

Suppose A has Rs x and B has Rs y

According to the given conditions,

x + 100 = 2(y − 100)
x + 100 = 2− 200
− 2y = −300                  ....(1)
and

y + 10 = 6(x − 10)
y + 10 = 6x − 60
6x − y = 70                      ....(2)

Multiplying equation (2) by 2 we get

12x − 2y = 140               ....(3)

Subtracting (1) from (3),  we get

11x = 440
x = 40

Substituting the value of x in equation (1), we get

40 − 2y = −300
−2y = −340
y = 170

Hence A has and B has

#### Page No 3.77:

Given:

(i) 5 pens and 6 pencils together cost of Rs. 9.

(ii) 3 pens and 2 pencils cost Rs. 5.

To Find: Cost of 1 pen and 1 pencil.

Let

(i) The cost of 1 pen = Rs x.

(ii) The cost of 1 pencil = Rs y.

According to question

Thus we get the following system of linear equation

By using cross multiplication we have

Cost of one pen =

Cost of one pencil =

#### Page No 3.77:

Given:

(i) 7 Audio cassettes and 3 Video cassettes cost is 1110.

(ii) 5 Audio cassettes and 4 Video cassettes cost Rs. 1350.

To Find: Cost of 1 audio cassette and 1 video cassettes.

Let (i) the cost of 1 audio cassette = Rs. x.

(ii) the cost of 1 video cassette = Rs. y.

According to the given conditions, we have

Thus, we get the following system of linear equation,

By using cross multiplication, we have

Hence cost of 1 audio cassette =

Hence cost of 1 video cassette =

#### Page No 3.77:

Given:

(i) Total numbers of pens and pencils = 40.

(ii) If she has 5 more pencil and 5 less pens, the number of pencils would be 4 times the number of pen.

To find: Original number of pens and pencils.

Suppose original number of pencil = x

And original number of pen = y

According the given conditions, we have,

Thus we got the following system of linear equations

Substituting the value of y from equation 1 in equation 2 we get

Substituting the value of y in equation 1 we get

Hence we got the result number of pencils is and number of pens are

#### Page No 3.77:

Given:

(i) Cost of 4 tables and 3 chairs = Rs 2250.

(ii) Cost of 3 tables and 4 chairs = Rs 1950.

To find: The cost of 2 chairs and 1 table.

Suppose, the cost of 1 table = Rs x.

The cost of 1 chair = Rs y.

According to the given conditions,

4x + 3y = 2250,

4x + 3y − 2200 = 0 …… (1)

3x + 4y = 1950,

3x + 4y − 1950 = 0 …… (2)

Solving eq. (1) and Eq. (2) by cross multiplication

∴ cost of 1 chairs = Rs. 150.

Hence total cost of 2 chairs and 1 table =

#### Page No 3.77:

Given:

(i) Cost of 3 bags and 4 pens = Rs. 257.

(ii) Cost of 4 bags and 3 pens = Rs. 324.

To Find: Cost of 1 bag and 10 pens.

Suppose, the cost of 1 bag = Rs. x.

and the cost 1 pen = Rs. y.

According to the given conditions, we have

3x + 4y = 257,

3x + 4y − 257 = 0 …… (1)

4x + 3y = 324

4x +3y − 324 = 0 …… (2)

Solving equation 1 and 2 by cross multiplication

Total cost of 1 bag and 10 pens =

Hence total cost of 1 bag and 10 pens =

#### Page No 3.77:

Given:

(i) Cost of 5 books and 7 pens = Rs. 79.

(ii) Cost of 7 books and 5 pens = Rs. 77.

To find: Cost of 1 book and 2 pens.

Suppose the cost of 1 book = Rs x.

and the cost of 1 pen = Rs y.

According to the given conditions, we have

5x + 7y = 79

5x + 7y − 79 = 0 …… (1)

7x + 5y = 77,

5x + 7y − 77 = 0 …… (2)

Thus we get the following system of linear equation,

Hence, the cost of 1 book = Rs 6

and the cost of 1 pen = Rs 7.

Therefore the cost of 2 pen = Rs 14.

Total cost of 1 book and 2 pens = 14 + 6 = 20

Total cost of 1 book and 2 pens =

Hence total cost of 1 book and 2 pens =

#### Page No 3.77:

To find:

(1) Total mangoes of A.

(2) Total mangoes of B.

Suppose A has x mangoes and B has y mangoes,

According to the given conditions,

3x + 6y + 270 = 0 …… (3) and

5y = 310

y =

Hence A has 34 mangoes and B has 62 mangoes.

#### Page No 3.77:

Given:

(i) On selling of a T.V. at 5% gain and a fridge at 10% gain, shopkeeper gain Rs.2000.

(ii) Selling T.V. at 10% gain and fridge at 5% loss. He gains Rs. 1500.

To find: Actual price of T.V. and fridge.

According to the question:

Hence we got the pair of equations

1x + 2y − 40000 = 0 …… (1)

2x − 1y − 30000 = 0 …… (2)

Solving the equation by cross multiplication method;

Cost of T.V. =

Cost of fridge =

Hence the cost of T.V. is and that of fridge is .

#### Page No 3.77:

Given: (i) 7 bats and 6balls cost is Rs3800

(ii) 3 bats and 5balls cost is Rs1750

To find: Cost of 1 bat and 1 ball

Let (i) the cost of 1 bat = Rs. x.

(ii) the cost of 1 ball = Rs. y.

According to the given conditions, we have

Thus, we get the following system of linear equation,

7x + 6y − 3800 = 0 …… (1)

3x + 5y − 1750 = 0 …… (2)

By using cross multiplication, we have

Hence cost of 1 bat =

Hence cost of 1 ball =

#### Page No 3.77:

To find:

(1) the fixed charge

(2) The charge for each day

Let the fixed charge be Rs x

And the extra charge per day be Rs y.

According to the given conditions,

Subtracting equation 1 and 2 we get

Substituting the value of y in equation 1 we get

Hence the fixed charge is and the charge of each day

#### Page No 3.82:

Let the numbers are x and y. One of them must be greater than or equal to the other. Let us assume that x is greater than or equal to y.

The sum of the two numbers is 8. Thus, we have

The sum of the two numbers is four times their difference. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

Multiplying the first equation by 5 and then adding with the second equation, we have

Substituting the value of x in the first equation, we have

Hence, the numbers are 5 and 3.

#### Page No 3.82:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.

The sum of the digits of the number is 13. Thus, we have

After interchanging the digits, the number becomes.

The difference between the number obtained by interchanging the digits and the original number is 45. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

Adding the two equations, we have

Substituting the value of x in the first equation, we have

Hence, the number is.

#### Page No 3.82:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.

The sum of the digits of the number is 5. Thus, we have

After interchanging the digits, the number becomes.

The number obtained by interchanging the digits is greater by 9 from the original number. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

Adding the two equations, we have

Substituting the value of x in the first equation, we have

Hence, the number is.

#### Page No 3.82:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.

The sum of the digits of the number is 15. Thus, we have

After interchanging the digits, the number becomes.

The number obtained by interchanging the digits is exceeding by 9 from the original number. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

Adding the two equations, we have

Substituting the value of x in the first equation, we have

Hence, the number is.

#### Page No 3.83:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.

The two digits of the number are differing by 2. Thus, we have

After interchanging the digits, the number becomes.

The sum of the numbers obtained by interchanging the digits and the original number is 66. Thus, we have

So, we have two systems of simultaneous equations

Here x and y are unknowns. We have to solve the above systems of equations for x and y.

(i) First, we solve the system

Adding the two equations, we have

Substituting the value of x in the first equation, we have

Hence, the number is.

(ii) Now, we solve the system

Adding the two equations, we have

Substituting the value of x in the first equation, we have

Hence, the number is.

There are two such numbers.

#### Page No 3.83:

Let the numbers are x and y. One of them must be greater than or equal to the other. Let us assume that x is greater than or equal to y.

The sum of the two numbers is 1000. Thus, we have

The difference between the squares of the two numbers is 256000. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

Adding the two equations, we have

Substituting the value of x in the first equation, we have

Hence, the numbers are 628 and 372.

#### Page No 3.83:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.

The two digits of the number are differing by 3. Thus, we have

After interchanging the digits, the number becomes.

The sum of the numbers obtained by interchanging the digits and the original number is 99. Thus, we have

So, we have two systems of simultaneous equations

Here x and y are unknowns. We have to solve the above systems of equations for x and y.

(i) First, we solve the system

Adding the two equations, we have

Substituting the value of x in the first equation, we have

Hence, the number is.

(ii) Now, we solve the system

Adding the two equations, we have

Substituting the value of x in the first equation, we have

Hence, the number is.

Note that there are two such numbers.

#### Page No 3.83:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.

The number is 4 times the sum of the two digits. Thus, we have

After interchanging the digits, the number becomes.

If 18 is added to the number, the digits are reversed. Thus, we have

So, we have the systems of equations

Here x and y are unknowns. We have to solve the above systems of equations for x and y.

Subtracting the first equation from the second, we have

Substituting the value of y in the first equation, we have

Hence, the number is.

#### Page No 3.83:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.

The number is 3 more than 4 times the sum of the two digits. Thus, we have

After interchanging the digits, the number becomes.

If 18 is added to the number, the digits are reversed. Thus, we have

So, we have the systems of equations

Here x and y are unknowns. We have to solve the above systems of equations for x and y.

Subtracting the first equation from the second, we have

Substituting the value of y in the first equation, we have

Hence, the number is.

#### Page No 3.83:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.

The number is 4 more than 6 times the sum of the two digits. Thus, we have

After interchanging the digits, the number becomes.

If 18 is subtracted from the number, the digits are reversed. Thus, we have

So, we have the systems of equations

Here x and y are unknowns. We have to solve the above systems of equations for x and y.

Multiplying the second equation by 5 and then subtracting from the first, we have

Substituting the value of y in the second equation, we have

Hence, the number is.

#### Page No 3.83:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.

The number is 4 times the sum of the two digits. Thus, we have

After interchanging the digits, the number becomes.

The number is twice the product of the digits. Thus, we have

So, we have the systems of equations

Here x and y are unknowns. We have to solve the above systems of equations for x and y.

Substituting in the second equation, we get

Or

Substituting the value of y in the first equation, we have

Hence, the number is.

Note that the first pair of solution does not give a two digit number.

#### Page No 3.83:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.

The product of the two digits of the number is 20. Thus, we have

After interchanging the digits, the number becomes.

If 9 is added to the number, the digits interchange their places. Thus, we have

So, we have the systems of equations

Here x and y are unknowns. We have to solve the above systems of equations for x and y.

Substituting from the second equation to the first equation, we get

Or

Substituting the value of y in the second equation, we have

Hence, the number is.

Note that in the first pair of solution the values of x and y are both negative. But, the digits of the number can’t be negative. So, we must remove this pair.

#### Page No 3.83:

Let the numbers are x and y. One of them must be greater than or equal to the other. Let us assume that x is greater than or equal to y.

The difference between the two numbers is 26. Thus, we have

One of the two numbers is three times the other number. Here, we are assuming that x is greater than or equal to y. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

Substitutingfrom the second equation in the first equation, we get

Substituting the value of y in the first equation, we have

Hence, the numbers are 39 and 13.

#### Page No 3.83:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.

The sum of the two digits of the number is 9. Thus, we have

After interchanging the digits, the number becomes.

Also, 9 times the number is equal to twice the number obtained by reversing the order of the digits. Thus, we have

So, we have the systems of equations

Here x and y are unknowns. We have to solve the above systems of equations for x and y.

Substituting from the second equation to the first equation, we get

Substituting the value of y in the second equation, we have

Hence, the number is.

#### Page No 3.83:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.

The difference between the two digits of the number is 3. Thus, we have

After interchanging the digits, the number becomes.

Seven times the number is equal to four times the number obtained by reversing the order of the digits. Thus, we have

So, we have two systems of simultaneous equations

Here x and y are unknowns. We have to solve the above systems of equations for x and y.

(i) First, we solve the system

Multiplying the first equation by 2 and then subtracting from the second equation, we have

Substituting the value of x in the first equation, we have

Hence, the number is.

(ii) Now, we solve the system

Multiplying the first equation by 2 and then subtracting from the second equation, we have

Substituting the value of x in the first equation, we have

But, the digits of the number can’t be negative. Hence, the second case must be removed.

#### Page No 3.85:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is

The numerator of the fraction is 4 less the denominator. Thus, we have

If the numerator is decreased by 2 and denominator is increased by 1, then the denominator is 8 times the numerator. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

Subtracting the second equation from the first equation, we get

Substituting the value of x in the first equation, we have

Hence, the fraction is.

#### Page No 3.85:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is

If 2 is added to both numerator and the denominator, the fraction becomes. Thus, we have

If 3 is added to both numerator and the denominator, the fraction becomes. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the fraction is.

#### Page No 3.85:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is

If 1 is subtracted from both numerator and the denominator, the fraction becomes. Thus, we have

If 1 is added to both numerator and the denominator, the fraction becomes. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the fraction is.

#### Page No 3.85:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is

If 1 is added to the numerator and 1 is subtracted from the denominator, the fraction becomes. Thus, we have

If 1 is added to the denominator, the fraction becomes. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the fraction is.

#### Page No 3.86:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is

If the numerator is multiplied by 2 and the denominator is reduced by 5, the fraction becomes. Thus, we have

If the denominator is doubled and the numerator is increased by 8, the fraction becomes. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the fraction is.

#### Page No 3.86:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is

If 3 is added to the denominator and 2 is subtracted from the numerator, the fraction becomes. Thus, we have

If 6 is added to the numerator and the denominator is multiplied by 3, the fraction becomes. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the fraction is.

#### Page No 3.86:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is

The sum of the numerator and the denominator of the fraction is. Thus, we have

If the denominator is increased by 2, the fraction reduces to. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the fraction is.

#### Page No 3.86:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is

If 2 is added to the numerator of the fraction, it reduces to. Thus, we have

If 1 is subtracted from the denominator, the fraction reduces to. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the fraction is.

#### Page No 3.86:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is

The sum of the numerator and denominator of the fraction is 4 more than twice the numerator. Thus, we have

If the numerator and denominator are increased by 3, they are in the ratio 2:3. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the fraction is.

#### Page No 3.86:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is

The sum of the numerator and denominator of the fraction is 3 less than twice the denominator. Thus, we have

If the numerator and denominator are decreased by 1, the numerator becomes half the denominator. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the fraction is.

#### Page No 3.86:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is

The sum of the numerator and denominator of the fraction is 12. Thus, we have

If the denominator is increased by 3, the fraction becomes. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the fraction is.

#### Page No 3.88:

Let the present age of father be x years and the present age of son be y years.

Father is three times as old as his son. Thus, we have

After 12 years, father’s age will beyears and son’s age will beyears. Thus using the given information, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the present age of father isyears and the present age of son isyears.

#### Page No 3.88:

Let the present age of A be x years and the present age of B be y years.

After 10 years, A’s age will beyears and B’s age will beyears. Thus using the given information, we have

Before 5 years, the age of A wasyears and the age of B wasyears. Thus using the given information, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the present age of A isyears and the present age of B isyears.

#### Page No 3.89:

Let the present ages of A, B, F and S be x, y, z and t years respectively.

A is elder to B by 2 years. Thus, we have

F is twice as old as A. Thus, we have

B is twice as old as S. Thus, we have

The ages of F and S is differing by 40 years. Thus, we have

So, we have four equations

……(1)

……(2)

……(3)

……(4)

Here x, y, z and t are unknowns. We have to find the value of x.

By using the third equation, the first equation becomes

From the fourth equation, we have

Hence, we have

Using the second equation, we have

Hence, the age of A isyears.

#### Page No 3.89:

Let the present age of the man be x years and the present age of his son be y years.

After 6 years, the man’s age will beyears and son’s age will beyears. Thus using the given information, we have

Before 3 years, the age of the man wasyears and the age of son’s wasyears. Thus using the given information, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the present age of the man isyears and the present age of son isyears.

#### Page No 3.89:

Let the present age of father be x years and the present age of his son be y years.

After 10 years, father’s age will beyears and son’s age will beyears. Thus using the given information, we have

Before 10 years, the age of father wasyears and the age of son wasyears. Thus using the given information, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the present age of father isyears and the present age of son isyears.

#### Page No 3.89:

Let the present age of father be x years and the present age of his son be y years.

The present age of father is three years more than three times the age of the son. Thus, we have

After 3 years, father’s age will beyears and son’s age will beyears.

Thus using the given information, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the present age of father isyears and the present age of son isyears.

#### Page No 3.89:

Let the present age of father be x years and the present age of his son be y years.

The present age of father is three times the age of the son. Thus, we have

After 12 years, father’s age will beyears and son’s age will beyears. Thus using the given information, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the present age of father isyears and the present age of son isyears.

#### Page No 3.89:

Let the present age of father be x years and the present ages of his two children’s be y and z years.

The present age of father is three times the sum of the ages of the two children’s. Thus, we have

After 5 years, father’s age will beyears and the children’s age will beandyears. Thus using the given information, we have

So, we have two equations

Here x, y and z are unknowns. We have to find the value of x.

Substituting the value offrom the first equation in the second equation, we have

By using cross-multiplication, we have

Hence, the present age of father isyears.

#### Page No 3.89:

Let the present age of father be x years and the present age of his son be y years.

After 2 years, father’s age will beyears and the age of son will beyears. Thus using the given information, we have

Before 2 years, the age of father wasyears and the age of son wasyears. Thus using the given information, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the present age of father isyears and the present age of son isyears.

#### Page No 3.89:

Let the present age of Nuri be x years and the present age of Sonu be y years.

After 10 years, Nuri’s age will be(x + 10) years and the age of Sonu will be(y + 10) years. Thus using the given information, we have

Before 5 years, the age of Nuri was(x – 5)years and the age of Sonu was(y – 5)years. Thus using the given information, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the present age of Nuri isyears and the present age of Sonu isyears.

#### Page No 3.89:

Let the present ages of Ani, Biju, Dharam and Cathy be x, y, z and t years respectively.

The ages of Ani and Biju differ by 3 years. Thus, we have

Dharam is twice as old as Ani. Thus, we have

Biju is twice as old as Cathy. Thus, we have

The ages of Cathy and Dharam differ by 30 years. Clearly, Dharam is older than Cathy. Thus, we have

So, we have two systems of simultaneous equations

(i)

(ii)

Here x, y, z and t are unknowns. We have to find the value of x and y.

(i) By using the third equation, the first equation becomes

From the fourth equation, we have

Hence, we have

Using the second equation, we have

From the first equation, we have

Hence, the age of Ani isyears and the age of Biju isyears.

(ii) By using the third equation, the first equation becomes

From the fourth equation, we have

Hence, we have

Using the second equation, we have

From the first equation, we have

Hence, the age of Ani isyears and the age of Biju isyears.

Note that there are two possibilities.

#### Page No 3.98:

We have to find the speed of car

Let and be two cars starting from pointsand respectively. Let the speed of car be x km/hr and that of carbe y km/hr.

Case I: When two cars move in the same directions:

Suppose two cars meet at point, Then,

Distance travelled by car

Distance travelled by car

It is given that two cars meet in 7 hours.

Therefore, Distance travelled by car X in hours = km

Distance traveled by car y in 7 hours = km

Clearly

Dividing both sides by common factor 7 we get,

Case II : When two cars move in opposite direction

Suppose two cars meet at point. Then,

Distance travelled by car,

Distance travelled by car.

In this case, two cars meet in 1 hour

Therefore Distance travelled by car X in1 hour km

Distance travelled by car Y in 1 hour km

From the above clearly,

...(ii)

By solving equation (i) and (ii), we get

Substituting in equation (ii) we get

Hence, the speed of car starting from point A is

The speed of car starting from point B is.

#### Page No 3.98:

Let the speed of the sailor in still water be x km/hr and the speed of the current be y km/hr

Speed upstream

Speed downstream

Now, Time taken to cover 8km down stream =

Time taken to cover 8km upstream=

But, time taken to cover 8 km downstream in 40 minutes or that is

Dividing both sides by common factor 2 we get

Time taken to cover 8km upstream in1hour

...(ii)

By solving these equation and we get

Substitute in equationwe get

Hence, the speed of sailor is

The speed of current is

#### Page No 3.98:

Let the speed of the boat in still water be x km/hr and the speed of the stream be y km/hr

Speed upstream =km/hr

Speed down stream =km/hr

Now,

Time taken to cover 30 km upstream = hrs

Time taken to cover 44 km down stream = hrs

But total time of journey is 10 hours

Time taken to cover 40 km upstream=

Time taken to cover 55 km down stream =

In this case total time of journey is given to be 13 hours

Therefore, ...(ii)

Putting = u and in equation and we get

Solving these equations by cross multiplication we get

and

Now,

By solving equation and we get ,

Substituting in equation we get ,

Hence, speed of the boat in still water is

Speed of the stream is

#### Page No 3.98:

We have to find the speed of the boat in still water and speed of the stream

Let the speed of the boat in still water be x km/hr and the speed of the stream be y km/hr then

Speed upstream

Sped down stream

Now, Time taken to cover km down stream =

Time taken to cover km upstream =

But, total time of journey is 6 hours

Time taken to cover km upstream =

Time taken to cover km down stream =

In this case total time of journey is given to or

...(ii)

By and in equation (i) and (ii) we get

Solving these equations by cross multiplication we get

and

and

Now,

and

By solving equation and we get,

By substituting in equation we get

Hence, the speed of the stream is

The speed of boat is

#### Page No 3.98:

Let the speed of Ajeet and Amit be x Km/hr respectively. Then,

Time taken by Ajeet to cover

Time taken by Amit to cover

By the given conditions, we have

If Ajeet doubles his speed, then speed of Ajeet is

Time taken by Ajeet to cover

Time taken by Amit to cover

According to the given condition, we have

...(ii)

Putting and, in equation (i) and (ii), we get

Adding equations (iii) and (iv), we get

Putting in equation (iii), we get

Now,

and

Hence, the speed of Ajeet is

The speed of Amit is

#### Page No 3.98:

Let the actual speed of the train be x Km/hr and the actual time taken by y hours. Then,

Distance covered=

If the speed is increased by, then time of journey is reduced by 1 hour i.e., when speed is , time of journey is

Distance covered= km

...(ii)

When the speed is reduced by, then the time of journey is increased by i.e., when speed is, time of journey is

Distance covered =

Thus we obtain the following equations

By using elimination method, we have

Putting the value in equation (iii) we get

Putting the value of x and y in equations (i) we get

Distance covered =

=

Hence, the distance is,

The speed of walking is .

#### Page No 3.98:

Let the speed of the train be x km/hour that of the car be y km/hr, we have the following cases

Case I: When Ramesh travels 760 Km by train and the rest by car

Time taken by Ramesh to travel 160 Km by train =

Time taken by Ramesh to travel (760-160) =600 Km by car =

Total time taken by Ramesh to cover 760Km = +

It is given that total time taken in 8 hours

Case II: When Ramesh travels 240Km by train and the rest by car

Time taken by Ramesh to travel 240 Km by train =

Time taken by Ramesh to travel (760-240) =520Km by car =

In this case total time of the journey is 8 hours 12 minutes

...(ii)

Putting and, , the equations and reduces to

Multiplying equation (iii) by 6 and (iv)by 20 the above system of equation becomes

Subtracting equation from we get

Putting in equation, we get

Now

and

Hence, the speed of the train is ,

The speed of the car is .

#### Page No 3.98:

Let the speed of the train be x km/hr that of the car be y km/hr, we have the following cases:

Case I: When a man travels 600Km by train and the rest by car

Time taken by a man to travel 400 Km by train =

Time taken by a man to travel (600-400) =200Km by car =hrs

Total time taken by a man to cover 600Km =

It is given that total time taken in 8 hours

Case II: When a man travels 200Km by train and the rest by car

Time taken by a man to travel 200 Km by train =

Time taken by a man to travel (600-200) = 400 Km by car

In this case, total time of the journey in 6 hours 30 minutes + 30 minutes that is 7 hours,

...(ii)

Putting and, , the equations and reduces to

Multiplying equation (iii) by 6 the above system of equation becomes

Substituting equation and, we get

Putting in equation, we get

Now

and

Hence, the speed of the train is,

The speed of the car is.

#### Page No 3.98:

Let x and y be two cars starting from points A and B respectively.

Let the speed of the car X be x km/hr and that of the car Y be y km/hr.

Case I: When two cars move in the same directions:

Suppose two cars meet at point Q, then,

Distance travelled by car X = AQ

Distance travelled by car Y = BQ

It is given that two cars meet in 8 hours.

Distance travelled by car X in 8 hours =km

AQ=

Distance travelled by car Y in 8 hours =km

BQ =

Clearly AQ-BQ = AB

Both sides divided by 8, we get

Case II: When two cars move in opposite direction

Suppose two cars meet at point P, then,

Distance travelled by X car X=AP

Distance travelled by Y car Y=BP

In this case, two cars meet in 1 hour 20 minutes, we can write it as 1 hour or

hours that is hours.

Therefore,

Distance travelled by car y in hours = km

Distance travelled by car y in hours = km

...(ii)

By solving (i) and (ii) we get,

By substituting in equation (ii), we get

Hence, speed of car X is , speed of car Y is.

#### Page No 3.98:

We have to find the speed of the boat in still water and speed of the stream

Let the speed of the boat in still water be km/hr and the speed of the stream be km/hr then

Speed upstream

Sped down stream

Now, Time taken to cover km upstream =

Time taken to cover km down stream =

But, total time of journey is 8 hours

Time taken to cover km upstream =

Time taken to cover km down stream =

In this case total time of journey is given to

...(ii)

By and in equation (i) and (ii) we get

Solving these equations by cross multiplication we get

and

and

Now,

and

By solving equation and we get,

By substituting in equation we get

Hence, the speed of boat in still water is ,

The speed of the stream is.

#### Page No 3.99:

Let the speed of the train be x km/hour that of the bus be y km/hr, we have the following cases

Case I: When Roohi travels 300 Km by train and the rest by bus

Time taken by Roohi to travel 60 Km by train =

Time taken by Roohi to travel (300-60) =240 Km by bus =

Total time taken by Roohi to cover 300Km= +

It is given that total time taken in 4 hours

Case II: When Roohi travels 100 km by train and the rest by bus

Time taken by Roohi to travel 100 Km by train =

Time taken by Roohi to travel (300-100) =200Km by bus =

In this case total time of the journey is 4 hours 10 minutes

...(i)

Putting and, , the equations and reduces to

Subtracting equation (iv) from (iii)we get

Putting in equation (iii), we get

Now

and

Hence, the speed of the train is,

The speed of the bus is.

#### Page No 3.99:

Let the speed of rowing in still water be x km/hr and the speed of the current be y km/hr

Speed upstream

Speed downstream

Now,

Time taken to cover km down stream =

Time taken to cover km upstream =

But, time taken to cover km downstream in

Time taken to cover km upstream in 2 hours

...(i)

By solving these equation (i) and (ii) we get

Substitute in equation (i)we get

Hence, the speed of rowing in still water is,

The speed of current is .4 km/ hr

#### Page No 3.99:

Let the speed of A and B be x Km/hr and y Km/hr respectively. Then,

Time taken by A to cover ,

And, Time taken by B to cover .

By the given conditions, we have

If A doubles his pace, then speed of A is

Time taken by A to cover ,

Time taken by B to cover .

According to the given condition, we have

Putting and, in equation (i) and (ii), we get

Adding equations (iii) and (iv), we get,

Putting in equation (iii), we get