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#### Question 1:

If in a rectangle, the length is increased and breadth reduced each by 2 units, the area is reduced by 28 square units. If, however the length is reduced by 1 unit and the breadth increased by 2 units, the area increases by 33 square units. Find the area of the rectangle.

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

#### Question 2:

The area of a rectangle remains the same if the length is increased by 7 meters and the breadth is decreased by 3 meters. The area remains unaffected if the length is decreased by 7 meters and breadth in increased by 5 meters. Find the dimensions of the rectangle.

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

#### Question 3:

In a rectangle, if the length is increased by 3 meters and breadth is decreased by 4 meters, the area of the rectangle is reduced by 67 square meters. If length is reduced by 1 meter and breadth is increased by 4 meters, the area is increased by 89 Sq. meters. Find the dimensions of the rectangle.

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.

#### Question 4:

The incomes of X and Y are in the ratio of 8 : 7 and their expenditures are in the ratio 19 : 16. If each saves Rs 1250, find their incomes.

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

#### Question 5:

A and B each has some money. If A gives Rs 30 to B, then B will have twice the money left with A. But, if B gives Rs 10 to A, then A will have thrice as much as is left with B. How much money does each have?

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

#### Question 6:

There are two examination rooms A and B. If 10 candidates are sent from A to B, the number of students in each room is same. If 20 candidates are sent from B to A, the number of students in A is double the number of students in B. Find the number of students in each room.

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.

#### Question 7:

2 men and 7 boys can do a piece of work in 4 days. The same work is done in 3 days by 4 men and 4 boys. How long would it take one man and one boy to do it?

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 .

#### Question 8:

In a ∆ABC, ∠A = x°, ∠B = (3x − 2)°, ∠C = y°. Also, ∠C − ∠B = 9°. Find the three angles.

Let , and

Substitute in above equation we get ,

${y}^{°}={7}^{°}+3{x}^{°}$

#### Question 9:

In a cyclic quadrilateral ABCD, ∠A = (2x + 4)°, ∠B = (y + 3)°, ∠C = (2y + 10)°, ∠D = (4x − 5)°. Find the four angles.

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 .

#### Question 10:

A railway half ticket costs half the full fare and the reservation charge is the same on half ticket as on full ticket. One reserved first class ticket from Mumbai to Ahmedabad costs Rs 216 and one full and one half reserved first class tickets cost Rs 327. What is the basic first class full fare and what is the reservation charge?

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 .

#### Question 11:

In a ∆ABC, ∠A = x°, ∠B = 3x° and ∠C = y°. If 3y − 5x = 30, prove that the triangle is right angled.

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

#### Question 12:

The car hire charges in a city comprise of a fixed charges together with the charge for the distance covered. For a journey of 12 km, the charge paid is Rs 89 and the journey of 20 km, the charge paid is Rs 145. What will a person have to pay for travelling a distance of 30 km?

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.

#### Question 13:

A part of monthly hostel charges in a college are fixed and the remaining depend on the number of days one has taken food in the mess. When a student A takes foods for 20 days, he has to pay Rs 1000 as hostel charges whereas a students B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charge and the cost of food per day.

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 .

#### Question 14:

Half the perimeter of a garden, whose length is 4 more than its width is 36 m. Find the dimension of the garden.

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

#### Question 15:

The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

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.

#### Question 16:

2 women and 5 men can together finish a piece of embroidery in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the embroidery, and that taken by 1 man alone.

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 .

#### Question 17:

A wizard having powers of mystic in candations and magical medicines seeing a cock, fight going on, spoke privately to both the owners of cocks. To one he said; if your bird wins, than you give me your stake-money, but if you do not win, I shall give you two third of that'. Going to the other, he promised in the same way to give three fourths. From both of them his gain would be only 12 gold coins. Find the the stake of money each of the cock-owners have.

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.

#### Question 18:

Meena went to a bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes Rs 50 and Rs 100 she received.

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

#### Question 19:

Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awanded for each correct answer and 2 marks been deducted for each incorrect answer, the Yash would have scored 50 marks. How many question were there in the test?

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 .

#### Question 20:

The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row there would be 2 rows more. Find the number of student in the class.

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.

#### Question 21:

One says. "give me hundred, friend! I shall then become twice as rich as you" The other replies, "If you give me ten, I shall be six times as rich as you". Tell me what is the amount of their respective capital?

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 .

#### Question 1:

Write the value of k for which the system of equations x + y − 4 = 0 and 2x + ky − 3 = 0 has no solution.

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.

#### Question 2:

Write the value of k for which the system of equations has infinitely many solutions.

$2x-y=5\phantom{\rule{0ex}{0ex}}6x+ky=15$

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.

#### Question 22:

ABCD is a cyclic quadrilateral such that ∠A = (4y + 20)°, ∠B = (3y − 5)°, ∠C = (−4x)° and ∠D = (7x + 5)°. Find the four angles.

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

#### Question 3:

Write the value of k for which the system of equations 3x − 2y = 0 and kx + 5y = 0 has infinitely may solutions.

The given equations are

For the equations to have infinite number of solutions,

Therefore,

By cross multiplication we have

Hence, the value of k for the system of equation and is.

#### Question 4:

Write the value of k for which the system of equations x + ky = 0, 2xy = 0 has unique solution.

The given equations are

For unique solution

For all real values of k, except the equations have unique solutions.

#### Question 5:

Write the set of values of a and b for which the following system of equations has infinitely many solutions.

$2x+3y=7\phantom{\rule{0ex}{0ex}}2ax+\left(a+b\right)y=28$

The given equations are

For the equations to have infinite number of solutions,

Therefore

Let us take

By dividing both the sides by 7 we get,

By multiplying equations by 2 we get

Substituting from we get

Subtracting in equation we have

Hence, the value of when system of equations has infinity many solutions.

#### Question 6:

For what value of k, the following pair of linear equation has infinitely many solutions?

$10x+5y-\left(k-5\right)=0\phantom{\rule{0ex}{0ex}}20x+10y-k=0$

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.

#### Question 7:

Write the number of solution of the following pair of linear equations:

x + 2y − 8 = 0
2x + 4y = 16

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.

#### Question 8:

Write the number of solutions of the following pair of linear equations:

x + 3y − 4 = 0
2x + 6y = 7

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.

#### Question 1:

The value of k for which the system of equations has a unique solution, is

kx y = 2
6x − 2y = 3

(a) =3
(b) ≠3
(c) ≠0
(d) =0

The given system of equations are

for unique solution

Here

By cross multiply we get

Hence, the correct choice is.

#### Question 2:

The value of k for which the system, of equations has infinite number of solutions, is

2x + 3y = 5
4x + ky = 10

(a) 1
(b) 3
(c) 6
(d) 0

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 .

#### Question 3:

The value of k for which the system of equations x + 2y − 3 = 0 and 5x + ky + 7 = 0 has no solution, is

(a) 10
(b) 6
(c) 3
(d) 1

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.

#### Question 4:

The value of k for which the system of equations 3x + 5y = 0 and kx + 10y = 0 has non-zero solution, is

(a) 0
(b) 2
(c) 6
(d) 8

The given system of equations are,

Here,

By cross multiply we get

Therefore the value of k is 6,

Hence, the correct choice is.

#### Question 5:

If the system of equations has infinitely many solutions, then

2x + 3y = 7
(a + b)x + (2ab)y = 21

(a) a = 1, b = 5
(b) a = 5, b = 1
(c) a = −1, b = 5
(d) a = 5, b = −1

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.

#### Question 6:

If the system of equations is inconsistent, then k =

$3x+y=1\phantom{\rule{0ex}{0ex}}\left(2k-1\right)x+\left(k-1\right)y=2k+1$

(a) 1
(b) 0
(c) −1
(d) 2

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 .

#### Question 7:

If ambl, then the system of equations

$ax+by=c\phantom{\rule{0ex}{0ex}}lx+my=n$

(a) has a unique solution
(b) has no solution
(c) has infinitely many solutions
(d) may or may not have a solution

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

#### Question 8:

If the system of equations has infinitely many solutions, then

$2x+3y=7\phantom{\rule{0ex}{0ex}}2ax+\left(a+b\right)y=28$

(a) a = 2b
(b) b = 2a
(c) a + 2b = 0
(d) 2a + b = 0

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 .

#### Question 9:

The value of k for which the system of equations has no solution is

$x+2y=5\phantom{\rule{0ex}{0ex}}3x+ky+15=0$

(a) 6
(b) −6
(c) 3/2
(d) None of these

The given system of equation is

If then the equation have no solution.

By cross multiply we get

Hence, the correct choice is.

#### Question 10:

If 2x − 3y = 7 and (a + b)x − (a + b − 3)y = 4a + b represent coincident lines, then a and b satisfy the equation

(a) a + 5b = 0
(b) 5a + b = 0
(c) a − 5b = 0
(d) 5ab = 0

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.

#### Question 11:

If a pair of linear equations in two variables is consistent, then the lines represented by two equations are

(a) intersecting
(b) parallel
(c) always coincident
(d) intersecting or coincident

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.

#### Question 12:

The area of the triangle formed by the line $\frac{x}{a}+\frac{y}{b}=1$ with the coordinate axes is

(a) ab
(b) 2ab
(c) $\frac{1}{2}ab$
(d) $\frac{1}{4}ab$

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 .

#### Question 13:

The area of the triangle formed by the lines y = x, x = 6 and y = 0 is

(a) 36 sq. units
(b) 18 sq. units
(c) 9 sq. units
(d) 72 sq. units

Given and

We have plotting points as when

Therefore, area of

Area of triangle is square units

Hence, the correct choice is .

#### Question 14:

If the system of equations 2x + 3y = 5, 4x + ky = 10 has infinitely many solutions, then k =

(a) 1
(b) ½
(c) 3
(d) 6

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 .

#### Question 15:

If the system of equations kx − 5y = 2, 6x + 2y = 7 has no solution, then k =

(a) −10
(b) −5
(c) −6
(d) −15

The given systems of equations are

If

Here

Hence, the correct choice is .

#### Question 16:

The area of the triangle formed by the lines x = 3, y = 4 and x = y is

(a) ½ sq. unit
(b) 1 sq. unit
(c) 2 sq. unit
(d) None of these

Given and

We have plotting points as when

Therefore, area of

Area of triangle is square units

Hence, the correct choice is

#### Question 17:

The area of the triangle formed by the lines 2x + 3y = 12, xy − 1 = 0 and x = 0 (as shown in Fig. 3.23), is

(a) 7 sq. units
(b) 7.5 sq. units
(c) 6.5 sq. units
(d) 6 sq. units

Given and

If We have plotting points as

Therefore, area of

Area of triangle is square units

Hence, the correct choice is

#### Question 1:

Akhila went to a fair in her village. She wanted to enjoy rides in the Giant Wheel and play Hoopla (a game in which you throw a rig on the items kept in the stall, and if the ring covers any object completely you get it.) The number of times she played Hoopla is half the number of rides she had on the Giant Wheel. Each ride costs Rs 3, and a game of Hoopla costs Rs 4. If she spent Rs 20 in the fair, represent this situation algebraically and graphically.

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.

#### Question 2:

Aftab tells his daughter, "Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be". Is not this interesting? Represent this situation algebraically and graphically.

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

#### Question 3:

The path of a train A is given by the equation 3x + 4y − 12 = 0 and the path of another train B is given by the equation 6x + 8y − 48 = 0. Represent this situation graphically.

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.

#### Question 4:

Gloria is walking along the path joining (−2, 3) and (2, −2), while Suresh is walking along the path joining (0, 5) and (4, 0). Represent this situation graphically.

Gloria is walking the path joining and

Suresh is walking the path joining and

 0 4 5 0

The graphical representations are

#### Question 5:

On comparing the ratios , and without drawing them, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincide :

(i) 5x − 4y + 8 = 0
7x + 6y − 9 = 0

(ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0

(iii) 6x − 3y + 10 = 0
2xy + 9 = 0

(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.

#### Question 6:

Given the linear equation 2x + 3y − 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is :

(i) intersecting lines
(ii) parallel lines
(iii) coincident lines

(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

#### Question 7:

The cost of 2kg of apples and 1 kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2kg of grapes is Rs. 300 Represent th situation algebraically and geometrically.

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.

#### Question 1:

Solve the following systems of equations graphically:

x + y = 3
2x + 5y = 12

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.

#### Question 2:

Solve the following systems of equations graphically:

x − 2y = 5
2x + 3y = 10

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

#### Question 3:

Solve the following systems of equations graphically:

3x + y + 1 = 0
2x − 3y + 8 = 0

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.

#### Question 4:

Solve the following systems of equations graphically:

2x + y − 3 = 0
2x − 3y − 7 = 0

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.

#### Question 5:

Solve the following systems of equations graphically:

x + y = 6
xy = 2

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.

#### Question 6:

Solve the following systems of equations graphically:

x − 2y = 6
3x − 6y = 0

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

#### Question 7:

Solve the following systems of equations graphically:

x + y = 4
2x − 3y = 3

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

#### Question 8:

Solve the following systems of equations graphically:

2x + 3y = 4
xy + 3 = 0

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.

#### Question 9:

Solve the following systems of equations graphically:

2x − 3y + 13 = 0
3x − 2y + 12 = 0

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.

#### Question 10:

Solve the following systems of equations graphically:

2x + 3y + 5 = 0
3x − 2y − 12 = 0

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.

#### Question 11:

Show graphically that each one of the following systems of equations has infinitely many solutions:

2x + 3y = 6
4x + 6y = 12

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.

#### Question 12:

Show graphically that each one of the following systems of equations has infinitely many solutions:

x − 2y = 5
3x − 6y = 15

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.

#### Question 13:

Show graphically that each one of the following systems of equations has infinitely many solutions:

3x + y = 8
6x + 2y = 16

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.

#### Question 14:

Show graphically that each one of the following systems of equations has infinitely many solutions:

x − 2y + 11 = 0
3x − 6y + 33 = 0

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.

#### Question 15:

Show graphically that each one of the following systems of equations is in-consistent (i.e. has no solution) :

3x − 5y = 20
6x − 10y = −40

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.

#### Question 16:

Show graphically that each one of the following systems of equations is in-consistent (i.e. has no solution) :

x − 2y = 6
3x − 6y = 0

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.

#### Question 17:

Show graphically that each one of the following systems of equations is in-consistent (i.e. has no solution) :

2yx = 9
6y − 3x = 21

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.

#### Question 18:

Show graphically that each one of the following systems of equations is in-consistent (i.e. has no solution) :

3x − 4y − 1 = 0
$2x-\frac{8}{3}y+5=0$

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.

#### Question 19:

Determine graphically the vertices of the triangle, the equations of whose sides are given below :

(i) 2yx = 8, 5yx = 14 and y − 2x = 1
(ii) y = x, y = 0 and 3x + 3y = 10

(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)

#### Question 20:

Determine, graphically whether the system of equations x − 2y = 2, 4x − 2y = 5 is consistent or in-consistent.

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

#### Question 21:

Determine, by drawing graphs, whether the following system of linear equations has a unique solution or not :

(i) 2x − 3y = 6, x + y = 1
(ii) 2y = 4x − 6, 2x = y + 3

(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

#### Question 22:

Solve graphically each of the following systems of linear equations. Also find the coordinates of the points where the lines meet axis of y.

(i) 2x − 5y + 4 = 0,
2x + y − 8 = 0

(ii) 3x + 2y = 12,
5x − 2y = 4

(iii) 2x + y − 11 = 0,
xy − 1 = 0

(iv) x + 2y − 7 = 0,
2xy − 4 = 0

(v) 3x + y − 5 = 0,
2xy − 5 = 0

(vi) 2xy − 5 = 0,
xy − 3 = 0

(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.

#### Question 23:

Determine graphically the coordinates of the vertices of a triangle, the equations of whose sides are :

(i) y = x, y = 2x and y + x = 6
(ii) y = x, 3y = x, x + y = 8

(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

#### Question 24:

Solve the following system of linear equation graphically and shade the region between the two lines and x-axis:

(i) 2x + 3y = 12,
xy = 1

(ii) 3x + 2y − 4 = 0,
2x − 3y − 7 = 0

(iii) 3x + 2y − 11 = 0
2x − 3y + 10 = 0

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.

#### Question 25:

Draw the graphs of the following equations on the same graph paper.

2x + 3y = 12,
xy = 1

Find the coordinates of the vertices of the triangle formed by the two straight lines and the y-axis.

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

#### Question 26:

Draw the graphs of xy + 1 = 0 and 3x + 2y − 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and x-axis and shade the triangular area. Calculate the area bounded by these lines and x-axis.

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 =

#### Question 27:

Solve graphically the system of linear equations:

4x − 3y + 4 = 0
4x + 3y − 20 = 0

Find the area bounded by these lines and x-axis.

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 =

#### Question 28:

Solve the following system of linear equations graphically :

3x + y − 11 = 0, x y − 1 = 0.
Shade the region bounded by these lines and y-axis. Also, find the area of the region bounded by the these lines and y-axis.

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

#### Question 29:

Solve graphically each of the following systems of linear equations. Also, find the coordinates of the points where the lines meet the axis of x in each system.

(i) 2x + y = 6
x − 2y = −2

(ii) 2xy = 2
4xy = 8

(iii) x + 2y = 5
2x − 3y = −4

(iv) 2x + 3y = 8
x − 2y = −3

(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.

#### Question 30:

Draw the graphs of the following equations:

2x − 3y + 6 = 0
2x + 3y − 18 = 0
y − 2 = 0

Find the vertices of the triangle so obtained. Also, find the area of the triangle.

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 =

#### Question 31:

Solve the following system of equations graphically.

2x − 3y + 6 = 0
2x + 3y − 18 = 0

Also, find the area of the region bounded by these two lines and y-axis.

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

#### Question 32:

Solve the following system of linear equations graphically.

4x − 5y − 20 = 0
3x + 5y − 15 = 0

Determine the vertices of the triangle formed by the lines representing the above equation and the y-axis.

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

#### Question 33:

Draw the graphs of the equations 5xy = 5 and 3xy = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and y-axis. Calculate the area of the triangle so formed.

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

#### Question 34:

Form the pair of linear equations in the following problems, and find their solution graphically:

(i) 10 students of class X took part in Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

(ii) 5 pencils and 7 pens together  cost Rs 50, whereas 7 pencils and 5 pens together cost Rs 46. Find the cost of one pencil and a pen.

(iii) Champa went to a 'sale' to purchase some pants and skirts. When her friends asked her how many of each she had bought, she answered, "The number of skirts is two less than twice the number of pants purchased. Also the number of skirts is four less than four times the number of pants purchased." Help her friends to find how many pants and skirts Champa bought.

(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.

#### Question 35:

Solve the following system of equations graphically:
Shade the region between the lines and the y-axis

(i) 3x − 4y = 7
5x + 2y = 3

(ii) 4xy = 4
3x + 2y = 14

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.

#### Question 36:

Represent the following pair of equations graphically and write the coordinates of points where the lines intersects y-axis.

x + 3y = 6
2x − 3y = 12

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.

#### Question 37:

Given the linear equation 2x + 3y − 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is

(i) intersecting lines
(ii) Parallel lines
(iii) coincident lines

(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−

#### Question 1:

Solve the following systems of equations:

$11x+15y+23=0\phantom{\rule{0ex}{0ex}}7x-2y-20=0$

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

#### Question 2:

Solve the following systems of equations:

3x − 7y + 10 = 0
y − 2x − 3 = 0

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

#### Question 3:

Solve the following systems of equations:

0.4x + 0.3y = 1.7
0.7x − 0.2y = 0.8

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

#### Question 4:

Solve the following systems of equations:

$\frac{x}{2}+y=0.8\phantom{\rule{0ex}{0ex}}\frac{7}{x+\frac{y}{2}}=10$

The given equations are:

Subtract (ii) from (i) we get

Put the value of in equation we get

Hence the value of and .

#### Question 5:

Solve the following systems of equations:

7(y + 3) − 2(x + 2) = 14
4(y − 2) + 3(x − 3) = 2

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

#### Question 6:

Solve the following systems of equations:

$\frac{x}{7}+\frac{y}{3}=5\phantom{\rule{0ex}{0ex}}\frac{x}{2}-\frac{y}{9}=6$

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 .

#### Question 7:

Solve the following systems of equations:

$\frac{x}{3}+\frac{y}{4}=11\phantom{\rule{0ex}{0ex}}\frac{5x}{6}-\frac{y}{3}=-7$

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 .

#### Question 8:

Solve the following systems of equations:

$\frac{4}{x}+3y=8\phantom{\rule{0ex}{0ex}}\frac{6}{x}-4y=-5$

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 .

#### Question 9:

Solve the following systems of equations:

$x+\frac{y}{2}=4\phantom{\rule{0ex}{0ex}}\frac{x}{3}+2y=5$

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

#### Question 10:

Solve the following systems of equations:

$x+2y=\frac{3}{2}\phantom{\rule{0ex}{0ex}}2x+y=\frac{3}{2}$

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

#### Question 11:

Solve the following systems of equations:

$\sqrt{2}x-\sqrt{3}y=0\phantom{\rule{0ex}{0ex}}\sqrt{3}x-\sqrt{8}y=0$

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

#### Question 12:

Solve the following systems of equations:

$3x-\frac{y+7}{11}+2=10\phantom{\rule{0ex}{0ex}}2y+\frac{x+11}{7}=10$

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

#### Question 13:

Solve the following systems of equations:

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

#### Question 14:

Solve the following systems of equations:

0.5x + 0.7y = 0.74
0.3x + 0.5y = 0.5

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

#### Question 15:

Solve the following systems of equations:

$\frac{1}{7x}+\frac{1}{6y}=3\phantom{\rule{0ex}{0ex}}\frac{1}{2x}-\frac{1}{3y}=5$

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

#### Question 16:

Solve the following systems of equations:

$\frac{1}{2x}+\frac{1}{3y}=2\phantom{\rule{0ex}{0ex}}\frac{1}{3x}+\frac{1}{2y}=\frac{13}{6}$

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 $y=\frac{1}{3}$

#### Question 17:

Solve the following systems of equations:

$\frac{x+y}{xy}=2\phantom{\rule{0ex}{0ex}}\frac{x-y}{xy}=6$

The given equations are:

Put the value of in equation, we get

Hence the value of and

#### Question 18:

Solve the following systems of equations:

$\frac{15}{u}+\frac{2}{\nu }=17\phantom{\rule{0ex}{0ex}}\frac{1}{u}+\frac{1}{\nu }=\frac{36}{5}$

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 .

#### Question 19:

Solve the following systems of equations:

$\frac{3}{x}-\frac{1}{y}=-9\phantom{\rule{0ex}{0ex}}\frac{2}{x}+\frac{3}{y}=5$

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

#### Question 20:

Solve the following systems of equations:

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 .

#### Question 21:

Solve the following systems of equations:

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

#### Question 22:

Solve the following systems of equations:

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

#### Question 23:

Solve the following systems of equations:

$\frac{6}{x+y}=\frac{7}{x-y}+3\phantom{\rule{0ex}{0ex}}\frac{1}{2\left(x+y\right)}=\frac{1}{3\left(x-y\right)\text{'}}$

where x + y ≠ 0 and xy ≠ 0

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.

#### Question 24:

Solve the following systems of equations:

$\frac{xy}{x+y}=\frac{6}{5}\phantom{\rule{0ex}{0ex}}\frac{xy}{y-x}=6,$

where x + y ≠ 0, yx ≠ 0

The given equations are:

Put the value of in equation, we get

Hence the value of and

#### Question 25:

Solve the following systems of equations:

$\frac{22}{x+y}+\frac{15}{x-y}=5\phantom{\rule{0ex}{0ex}}\frac{55}{x+y}+\frac{45}{x-y}=14$

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 .

#### Question 26:

Solve the following systems of equations:

$\frac{5}{x+y}-\frac{2}{x-y}=-1\phantom{\rule{0ex}{0ex}}\frac{15}{x+y}+\frac{7}{x-y}=10$

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 .

#### Question 27:

Solve the following systems of equations:

$\frac{3}{x+y}+\frac{2}{x-y}=2\phantom{\rule{0ex}{0ex}}\frac{9}{x+y}-\frac{4}{x-y}=1$

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 .

#### Question 28:

Solve the following systems of equations:

$\frac{1}{2\left(x+2y\right)}+\frac{5}{3\left(3x-2y\right)}=\frac{-3}{2}\phantom{\rule{0ex}{0ex}}\frac{5}{4\left(x+2y\right)}-\frac{3}{5\left(3x-2y\right)}=\frac{61}{60}$

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 .

#### Question 29:

Solve the following systems of equations:

$\frac{5}{x+1}-\frac{2}{y-1}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\frac{10}{x+1}+\frac{2}{y-1}=\frac{5}{2}$

where x ≠ −1 and y ≠ 1

The given equations are:

Let and then equations are

Put the value of in equation, we get

Then

Hence the value of and .

#### Question 30:

Solve the following systems of equations:

x + y = 5xy
3x + 2y = 13xy,

x ≠ 0, y ≠ 0

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

#### Question 31:

Solve the following systems of equations:

$x+y=2xy\phantom{\rule{0ex}{0ex}}\frac{x-y}{xy}=6$

x ≠ 0, y ≠ 0

The given equations are:

Put the value of in equation, we get

Hence the value of and .

#### Question 32:

Solve the following systems of equations:

2(3u − ν) = 5uν
2(u + 3ν) = 5uν

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 .

#### Question 33:

Solve the following systems of equations:

$\frac{2}{3x+2y}+\frac{3}{3x-2y}=\frac{17}{5}\phantom{\rule{0ex}{0ex}}\frac{5}{3x+2y}+\frac{1}{3x-2y}=2$

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 .

#### Question 34:

Solve the following systems of equations:

$\frac{4}{x}+3y=14\phantom{\rule{0ex}{0ex}}\frac{3}{x}-4y=23$

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 .

#### Question 35:

Solve the following systems of equations:

99x + 101y = 499
101x + 99y = 501

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

#### Question 36:

Solve the following systems of equations:

23x − 29y = 98
29x − 23y = 110

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

#### Question 37:

Solve the following systems of equations:

xy + z = 4
x − 2y − 2z = 9
2x + y + 3z = 1

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.

#### Question 38:

Solve the following systems of equations:

xy + z = 4
x + y + z = 2
2x + y − 3z = 0

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.

#### Question 39:

Solve the following systems of equations:

$\frac{44}{x+y}+\frac{30}{x-y}=10\phantom{\rule{0ex}{0ex}}\frac{55}{x+y}+\frac{40}{x-y}=13$

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

#### Question 40:

Solve the following systems of equations:

$\frac{4}{x}+5y=7\phantom{\rule{0ex}{0ex}}\frac{3}{x}+4y=5$

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

#### Question 41:

Solve the following systems of equations:

$\frac{2}{x}+\frac{3}{y}=13\phantom{\rule{0ex}{0ex}}\frac{5}{x}-\frac{4}{y}=-2$

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 .

#### Question 42:

Solve the following systems of equations:

$\frac{5}{x-1}+\frac{1}{y-2}=2\phantom{\rule{0ex}{0ex}}\frac{6}{x-1}-\frac{3}{y-2}=1$

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 .

#### Question 43:

Solve the following systems of equations:

$\frac{10}{x+y}+\frac{2}{x-y}=4\phantom{\rule{0ex}{0ex}}\frac{15}{x+y}-\frac{9}{x-y}=-2$

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 .

#### Question 44:

Solve the following systems of equations:

$\frac{1}{3x+y}+\frac{1}{3x-y}=\frac{3}{4}\phantom{\rule{0ex}{0ex}}\frac{1}{2\left(3x+y\right)}-\frac{1}{2\left(3x-y\right)}=-\frac{1}{8}$

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

#### Question 45:

Solve the following systems of equations:

$\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2\phantom{\rule{0ex}{0ex}}\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1$

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

#### Question 46:

Solve the following systems of equations:

$\frac{7x-2y}{xy}=5\phantom{\rule{0ex}{0ex}}\frac{8x+7y}{xy}=15$

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

#### Question 47:

Solve the following systems of equations:

152x − 378y = −74
−378x + 152y = −604

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

#### Question 1:

Solve each of the following systems of equations by the method of cross-multiplication :

x + 2y + 1 = 0
2x − 3y − 12 = 0

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

#### Question 2:

Solve each of the following systems of equations by the method of cross-multiplication :

3x + 2y + 25 = 0
2x + y + 10 = 0

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

#### Question 3:

Solve each of the following systems of equations by the method of cross-multiplication :

2x + y = 35
3x + 4y = 65

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

#### Question 4:

Solve each of the following systems of equations by the method of cross-multiplication :

2xy = 6
xy = 2

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

#### Question 5:

Solve each of the following systems of equations by the method of cross-multiplication :

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

#### Question 6:

Solve each of the following systems of equations by the method of cross-multiplication :

ax + by = ab
bxay = a + b

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

#### Question 7:

Solve each of the following systems of equations by the method of cross-multiplication :

x + ay = b
axby = c

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

#### Question 8:

Solve each of the following systems of equations by the method of cross-multiplication :

ax + by = a2
bx + ay = b2

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

#### Question 9:

Solve each of the following systems of equations by the method of cross-multiplication :

$\frac{x}{a}+\frac{y}{b}=2\phantom{\rule{0ex}{0ex}}ax-by={a}^{2}-{b}^{2}$

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

#### Question 10:

Solve each of the following systems of equations by the method of cross-multiplication :

$\frac{x}{a}+\frac{y}{b}=a+b\phantom{\rule{0ex}{0ex}}\frac{x}{{a}^{2}}+\frac{y}{{b}^{2}}=2$

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

#### Question 11:

Solve each of the following systems of equations by the method of cross-multiplication :

$\frac{x}{a}=\frac{y}{b}\phantom{\rule{0ex}{0ex}}ax+by={a}^{2}+{b}^{2}$

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

#### Question 12:

Solve each of the following systems of equations by the method of cross-multiplication :

$\frac{5}{x+y}-\frac{2}{x-y}=-1\phantom{\rule{0ex}{0ex}}\frac{15}{x+y}+\frac{7}{x-y}10,$

where x ≠ 0 and y ≠ 0

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

#### Question 13:

Solve each of the following systems of equations by the method of cross-multiplication :

$\frac{2}{x}+\frac{3}{y}=13\phantom{\rule{0ex}{0ex}}\frac{5}{x}-\frac{4}{y}=-2$

where x ≠ 0 and y ≠ 0

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

#### Question 14:

Solve each of the following systems of equations by the method of cross-multiplication :

$ax+by=\frac{a+b}{2}\phantom{\rule{0ex}{0ex}}3x+5y=4$

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

#### Question 15:

Solve each of the following systems of equations by the method of cross-multiplication :

2ax + 3by = a + 2b
3ax + 2by = 2a + b

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

#### Question 16:

Solve each of the following systems of equations by the method of cross-multiplication :

5ax + 6by = 28
3ax + 4by = 18

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

#### Question 17:

Solve each of the following systems of equations by the method of cross-multiplication :

(a + 2b)x + (2ab)y = 2
(a − 2b)x + (2a + b)y = 3

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

#### Question 18:

Solve each of the following systems of equations by the method of cross-multiplication :

$x\left(a-b+\frac{ab}{a-b}\right)=y\left(a+b-\frac{ab}{a+b}\right)\phantom{\rule{0ex}{0ex}}x+y=2{a}^{2}$

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

#### Question 19:

Solve each of the following systems of equations by the method of cross-multiplication :

$bx+cy=a+b\phantom{\rule{0ex}{0ex}}ax\left(\frac{1}{a-b}-\frac{1}{a+b}\right)+cy\left(\frac{1}{b-a}-\frac{1}{b+a}\right)=\frac{2a}{a+b}$

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

#### Question 20:

Solve each of the following systems of equations by the method of cross-multiplication :

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

#### Question 21:

Solve each of the following systems of equations by the method of cross-multiplication :

${a}^{2}x+{b}^{2}y={c}^{2}\phantom{\rule{0ex}{0ex}}{b}^{2}x+{a}^{2}y={d}^{2}$

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

#### Question 22:

Solve each of the following systems of equations by the method of cross-multiplication :

$\frac{57}{x+y}+\frac{6}{x-y}=5\phantom{\rule{0ex}{0ex}}\frac{38}{x+y}+\frac{21}{x-y}=9$

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

#### Question 23:

Solve each of the following systems of equations by the method of cross-multiplication :

$2\left(ax-by\right)+a+4b=0\phantom{\rule{0ex}{0ex}}2\left(bx+ay\right)+b-4a=0$

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

#### Question 24:

Solve each of the following systems of equations by the method of cross-multiplication :

$6\left(ax+by\right)=3a+2b\phantom{\rule{0ex}{0ex}}6\left(bx-ay\right)=3b-2a$

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

#### Question 25:

Solve each of the following systems of equations by the method of cross-multiplication :

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

#### Question 26:

Solve each of the following systems of equations by the method of cross-multiplication :

$mx-ny={m}^{2}+{n}^{2}\phantom{\rule{0ex}{0ex}}x+y=2m$

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

#### Question 27:

Solve each of the following systems of equations by the method of cross-multiplication :

$\frac{ax}{b}-\frac{by}{a}=a+b\phantom{\rule{0ex}{0ex}}ax-by=2ab$

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

#### Question 28:

Solve each of the following systems of equations by the method of cross-multiplication :

$\frac{b}{a}x+\frac{a}{b}y={a}^{2}+{b}^{2}\phantom{\rule{0ex}{0ex}}x+y=2ab$