Rs Aggarwal 2017 Solutions for Class 9 Math Chapter 8 Linear Equations In Two Variables are provided here with simple step-by-step explanations. These solutions for Linear Equations In Two Variables are extremely popular among Class 9 students for Math Linear Equations In Two Variables Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2017 Book of Class 9 Math Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2017 Solutions. All Rs Aggarwal 2017 Solutions for class Class 9 Math are prepared by experts and are 100% accurate.

#### Question 1:

Draw the graph of each of the following equations:
(i) x = 5
(ii) y = −2
(iii) x + 6 = 0
(iv) x + 7 = 0
(v) y = 0
(vi) x = 0

(i) x = 5
(ii) y = -2
(iii) x + 6 = 0 ⇒ x = -6
(iv) x + 7 = 0 ⇒ x = -7
(v) y = 0
(vi) x = 0

#### Question 2:

Draw the graph of the equation y = 3x. From your graph, find the value of y when x = −2.

Given equation:  y = 3x

When x = -2, y = -6.
When x = -1, y = -3.
Thus, we have the following table:

 x -2 -1 y -6 -3

Now plot the points (-2,-6), (-1, -3) on a graph paper.
Join the points and extend the line in both the directions.
The line segment is the required  graph of the equation.

Also, when x = -2, y = -6.

#### Question 3:

Draw the graph of the equation x + 2y − 3 = 0. From your graph, find the value of y when x = 5.

Given equation: x + 2y - 3 = 0
Or,  x + 2y = 3
When y = 0, x + 0 = 3 ⇒ x = 3
When y = 1, x + 2 = 3 ⇒ x = 3-2 = 1
When y = 2, x + 4 = 3 ⇒ x = 3 - 4 = -1
Thus, we have the following table:

 x 3 1 -1 y 0 1 2

Now plot the points (3,0) ,(1,1) and (-1,2) on the graph paper.
Join the points and extend the line in both the directions.
The line segment is the required graph of the equation.

When x = 5,

#### Question 4:

Draw the graph of each of the following equations:
(i) y = x
(ii) y = x
(iii) y + 3x = 0
(iv) 2x + 3y = 0
(v) 3x − 2y = 0
(vi) 2x + y = 0

(i) Given equation: y = x
When x = 0, y = 0.
When x = 1, y = 1.
Thus, we have the following table:

 x 0 1 y 0 1

Now, plot the points (0,0) and (1,1).
Join the points and extend the line in both the directions.
The line segment is the required graph of y = x.

(ii) Given equation: y = -x
When x = 0, y = 0.
When x = 1, y = -1.
Thus, we have the following table:
 x 0 1 y 0 -1

Now, plot the points (0,0) and (1,-1).
Join the points and extend the line in both the directions.
The line segment is the required graph of y = -x.

(iii) Given equation: y + 3x = 0
When x = 0, y = 0.
When x = 1, y = -3.
Thus, we have the following table:
 x 0 1 y 0 -3

Now, plot the points (0,0) and (1,-3)
Join the points and extend the line in both the directions.
Thus, the line segment is the required graph of y + 3x = 0.

(iv) Given equation: 2x+ 3y = 0
When x = 0, y = 0.
When x = 3, y = -2.
Thus, we have the following table:
 x 0 3 y 0 -2

Now, plot the points (0,0) and (3,-2).
Join the points and extend the line in both the directions.
The line segment is the required graph of 2x+ 3y = 0

(v) Given equation: 3x - 2y = 0
When x = 0, y = 0.
When x = 2, y = 3.
Thus, we have the following table:
 x 0 2 y 0 3

Now, plot the points (0,0) and (2,3).
Join the points and extend the line in both the directions.
The line segment is the required graph of 3x - 2y = 0

(vi) Given equation: 2x + y = 0
When x = 0, y = 0.
When x = 1, y = -2.
Thus, we have the following table:
 x 0 1 y 0 -2

Now, plot the points (0,0) and (1,-2)
Join the points and extend the line in both the directions.
The line segment is the required graph of 2x + y = 0.

#### Question 5:

Draw the graph of the equation 2x − 3y = 5. From the graph, find (i) the value of y when x = 4 and (ii) the value of x when y = 3.

Given equation :

When,

When,
Thus, we have the following table:

 $\text{x}$ 1 -2 $y$ -1 -3
Plot the points  on the graph paper and extend the line in both directions.

(i) When x = 4:

(ii) When y = 3:

#### Question 6:

Draw the graph of the equation 2x + y = 6. Find the coordinates of the point, where the graph cuts the x-axis.

Given equation:

When, .
When, .
When, .
Thus, we have the following table:

 $\text{x}$ 0 1 2 $y$ 6 4 2

Plot the points  and $\left(2,2\right)$ on the graph paper. Join these points and extend the line.

Clearly, the graph cuts the $x-axis$ at $P\left(3,0\right)$.

#### Question 7:

Draw the graph of the equation 3x + 2y = 6. Find the coordinates of the point, where the graph cuts the y-axis.

Given equation: . Then,

When ,

When ,
Thus, we get the following table:

 $\text{x}$ 2 4 $y$ 0 -3

Plot the points  on the graph paper. Join the points and extend the graph in both the directions.

Clearly, the graph cuts the $y-axis$ at P(0,3).

#### Question 1:

x = 0 is the equation of
(a) the x-axis
(b) the y-axis
(c) a line parallel to the x-axis
(d) a line parallel to the y-axis

(b) the y-axis
x = 0 is the equation of the y-axis.
So, the graph of x = 0 is YOY​'.

#### Question 2:

y = 0 is the equation of
(a) the x-axis
(b) the y-axis
(c) a line parallel to the x-axis
(d) a line parallel to the y-axis

(a) the x-axis
y = 0 is the equation of the x- axis.
So, the graph of y = 0 is X'OX.

#### Question 3:

x + 3 = 0 is the equation of a line
(a) parallel to the x-axis and passing through (−3, 0)
(b) parallel to the y-axis and passing through (−3, 0)
(c) parallel to the y-axis and passing through (0, −3)
(d) None of these

(b) parallel to the y-axis and passing through (−3, 0)
Given equation:
Or,
The equation of the line is parallel to the $y$-axis and passes through $\left(-3,0\right)$.

#### Question 4:

y − 4 = 0 is the equation of a line
(a) parallel to the x-axis and passing through (4, 0)
(b) parallel to the x-axis and passing through (0, 4)
(c) parallel to the y-axis and passing through (0, 4)
(d) None of these

(b) parallel to the x-axis and passing through (0, 4)
Given equation:
Or ,
The equation of the line is parallel to the $\text{x}$-axis and passes through $\left(0,4\right)$.

#### Question 5:

The point of the form (a, a), where a ≠ 0, lies on
(a) the x-axis
(b) the y-axis
(c) the line y = x
(d) the line x + y = 0

(c) the line y = x
Given, a point of the form , where $a$≠ 0 .
When , the point is (1,1)
When , the point is (2,2).....and so on.
Plot the points (1,1) and (2,2).......and so on. Join the points and extend them in both the direction. You will get the equation of the line .

#### Question 6:

The point of the form (a, a), where a ≠ 0, lies on
(a) the x-axis
(b) the y-axis
(c) the line y = x
(d) the line x + y = 0

(d) the line x + y = 0
Given, a point of the form , where $a$ ≠ 0.
When , the point is (1,-1).
When , the point is (2,-2).
When , the point is (3,-3).......and so on.
Plot these points on a graph paper. Join these points and extend them in both the directions.
You will get the equation of the line .

#### Question 7:

The linear equation 3x − 5y = has
(a) a unique solution
(b) two solutions
(c) infinitely many solutions
(d) no solution

(c) infinitely many solutions
Given linear equation:
Or,

When , .

When  , .

When , .

Thus, we have the following table:

 $\text{x}$ 5 10 0 $y$ 0 3 -3

Plot the points  and $C\left(0,-3\right)$. Join the points and extend them in both the directions.
We get infinite points that satisfy the given equation.
Hence, the linear equation has infinitely many solutions.

#### Question 8:

The graph of the linear equation 3x + 2y = 6 cuts the y-axis at the point
(a) (2, 0)
(b) (0, 2)
(c) (0, 3)
(d) (3, 0)

(c) (0, 3)
Given linear equation:
Or,

Putting we get:

Thus, the graph cuts the y-axis at (0,3).

#### Question 9:

The graph of the linear equation 4x + 3y = 12 cuts the y-axis at the point
(a) (4, 0)
(b) (0, 4)
(c) (0, 3)
(d) (3, 0)

(d) (3, 0)
Given equation:

When, y = 0, x =
Thus, the graph cuts the x-axis at (3,0).

#### Question 10:

The graph of the line x = 3 passes through the point
(a) (0, 3)
(b) (2, 3)
(c) (3, 2)
(d) None of these

(c) (3, 2)
The graph of line x = 3 is a line parallel to the y-axis.
Hence, its passes through (3,2), satisfying x =3.

#### Question 11:

The graph of the line y = 2 passes through the point
(a) (2, 0)
(b) (2, 3)
(c) (5, 2)
(d) None of these

(c) (5, 2)
The graph of the line y = 2 is a line parallel to the x-axis.
Thus, it passes through (5,2), satisfying y =2.

#### Question 12:

The graph of the line y = −3 does not pass through the point
(a) (2, −3)
(b) (3, −3)
(c) (0, −3)
(d) (−3, 2)

(d) (−3, 2)

The graph of the line y = -3 does not pass through (-3,2) since (-3,2) does not satisfy y = -3.

#### Question 13:

A linear equation in two variables x and y is of the form ax + by + c = 0, where
(a) a ≠ 0, b ≠ 0
(b) a ≠ 0, b = 0
(c) a = 0, b ≠ 0
(d) a = 0, c = 0

(a) a ≠ 0, b ≠ 0
A linear equation in two variables $\text{x}$ and $y$ is of the form , where $a$ ≠ 0 and $b$ ≠ 0.

#### Question 14:

Any point on the x-axis is of the form
(a) (x, y), where x ≠ 0 and y ≠ 0
(b) (0, y), where y ≠ 0
(c) (x, 0), where x ≠ 0
(d) (y, y), where y ≠ 0

(c) (x, 0), where x ≠ 0
Any point on the $\text{x}$-axis is of the form $\left(\text{x,0)}$, where $\text{x}$ ≠ 0.

#### Question 15:

Any point on the y-axis is of the form:
(a) (x, 0), where x ≠ 0
(b) (0, y), where y ≠ 0
(c) (x, x), where x ≠ 0
(d) None of these

(b) (0, y), where y ≠ 0
Any point on the $y$-axis is of the form $\left(0,y\right)$, where $y$ ≠ 0.

#### Question 16:

How many linear equations in x and y can be satisfied by x = 2, y = 3?
(a) Only one
(b) Only two
(c) Infinitely many
(d) None of these

(c) Infinitely many
Infinite linear equations are satisfied by .

#### Question 17:

The graph of the linear equation 3x + 2y = 6 is the line which meets the x-axis at the point
(a) (0, 3)
(b) (2, 0)
(c) (2, 3)
(d) (3, 2)

(b) (2, 0)
Given equation:
Or,

When y = 0, .
Thus, the graph meets the x-axis at (2,0).

#### Question 18:

The graph of the linear equation 2x + 5y = 10 is the line which meets the y-axis at the point
(a) (0, 2)
(b) (5, 0)
(c)
(d) (2, 1.2)

(a) (0, 2)
Given equation:

When ,
Thus, the graph meets the y-axis at (0,2).

#### Question 19:

If each of (2, 2), (0, 0) and (2, −2) is a solution of a linear equation in x and y, the equation is
(a) x y = 0
(b) x + y = 0
(c) −2x + y = 0
(d) −x + 2y = 0

(b) x + y = 0
Given points:  and
We have to check which equation satisfies the given points.
Let us check for (a)
Substituting in the equation, we get:
Substituting  in the equation, we get:
Substituting  in the equation, we get:
So, the given points do not satisfy the equation.
Now, let us check (b)
Substituting  in the equation, we get:
Substituting  in the equation, we get:
Substituting $\left(2,-2\right)$ in the equation, we get:
So, the given points satisfy the equation.
Similarly, we can check other equations and find that the given points will not satisfy the equations.
Hence, each of  and  is a solution of the linear equation .

#### Question 20:

The graph of the linear equation x y = 0 passes through the point
(a)
(b)
(c) (0, −1)
(d) (1, 1)

(d) (1, 1)
Given equation: or,
When ,
When , ... and so on
Thus, we get the following table:

 $\text{x}$ 1 2 $y$ 1 2

​Plot the points on the graph paper. Join the points and extend them in both the directions.
We can see the linear equation  passes through the point (1,1).

#### Question 21:

Assertion: x = 3 is a line parallel to the y-axis.
Reason: The equation of a line parallel to the y-axis is x = a.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
We know that equation of a line parallel to the $y$-axis is . So, Reason (R) is true.
Also,  is a line parallel to the $y$-axis. So, Assertion (A) is true.
Thus, Reason (R) and Assertion (A) are true and Reason (R) is a correct explanation of Assertion (A).

#### Question 22:

Assertion: y = mx represents a line passing through the origin.
Reason: Any line parallel to x-axis is y = b.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
We know that  is the equation of a line passing through the origin.

To check we will put the value of origin(0,0) in the equation:
LHS: y = 0
RHS : m(0)  = 0
Since, LHS  = RHS = 0
Therefore, origin satisfies the given equation.

Moreover, we know that  is the equation of the line parallel to the $\text{x}$-axis.
So, Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).

#### Question 23:

Assertion: x + y = 5 is the equation of a line passing through the origin.
Reason: y = mx is the equation of a line passing through the origin.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

(d) Assertion is false and Reason is true.
Given equation:
Or,
When

Thus, we get the following table:

 $\text{x}$ 1 2 3 4 5 $y$ 4 3 2 1 0

Plot the points  and . Join these points and extend them in both the directions.
We get a line that does not pass through the origin.
Moreover, we know that  is the equation of the line passing through the origin.
So, Assertion (A) is false and Reason (R) is true.

#### Question 24:

Match the following columns:

 Column I Column II (a) The equation of a line parallel to the x-axis is (p) y = mx (b) The equation of a line parallel to the y-axis is (q) $\frac{5}{2}$ (c) The equation of a line through the origin is (r) x = k (d) If the point (2, 3) lies on the graph of the equation 3y = ax + 4, then a = (s) y = k
(a) ......,
(b) ......,
(c) ......,
(d) ......,

(a) The equation of a line parallel to the $x-$axis is .
(b)The equation of a line parallel to the $y$-axis is .
(c) The equation of a line through the origin is .
(d) The point (2, 3) lies on the graph of . Then,

 Column I Column II (a) The equation of a line parallel to the x- axis is (s) y = k (b) The equation of a line parallel to the y- axis is (r) x = k (c) The equation of a line through the origin is (p) y = mx (d) If the point (2, 3) lies on the graph of the equation  3y = ax + 4, then a = (q) $\frac{5}{2}$

#### Question 25:

Write each of the following in the form ax + by + c = 0.
(i) x = −2
(ii) y = 6

(i)

(ii)

#### Question 26:

Write each of the following in the form ax + by + c = 0.
(i) 3x = 5
(ii) 5y = 4

(i)   or

(ii)   or

#### Question 27:

The total runs scored by two batsmen in a one-day cricket match is 215. Express this information in the form of a linear equation in two variables.

Let $\text{x}$ be the number of  runs scored by the first batsman and $y$ be the number of runs scored by the second batsman.
Then, total runs =  215    (Given)
So, the linear equation in two variable is .

#### Question 28:

The weight of a book is three times the weight of a note book. Express this fact in the form of an equation in two variables.

Let the weight of notebook be $\text{x}$ gm.
Let the weight of book be $y$ gm.
According to the given condition, we have:

Hence, the equation in two variables is .

#### Question 29:

Check which of the following are solutions of the equation 2x 3y = 6.
(i) (3, 0)
(ii) (0, 2)
(iii) (2, 6)
(iv) (6, 2)

(i)
Given equation:
Putting  in the given equation, we get:

So, it satisfies the given equation.

(ii)
Given equation:
Putting  in the given equation, we get:
≠ $RHS$
So, it does not satisfy the given equation.

(iii)
Given equation:
Putting  in the given equation, we get:
≠ $RHS$
So, it does not satisfy the given equation.

(iv)
Given equation:
Putting  in the given equation, we get:

So, it satisfies the given equation.

Hence,  are the solutions of the given equation.

#### Question 30:

Find the value of k if x = 3, y = 1 is a solution of the equation 2x + 5y = k.

Given equation: ;  are the solutions of the given equation.
So, they must satisfy the given equation.
Putting  in the equation, we get:

Hence, the value of .

#### Question 31:

Find four different solutions of 2x + y = 6.

Given:
Or,
When , .
When , .
When , .
When , .
Hence, four different solutions are  and $\left(4,-2\right)$.

#### Question 32:

Express y in terms of x, given that $\frac{x}{5}+2y=3$. Check whether (−5, 2) is a solution of the given equation.

Given:

We have to check whether  is a solution of the given equation.
So,

Hence,  is a solution of the given equation.

#### Question 33:

Show that (3, 1) as well as (2, −2) are the solutions of the equation 3x y = 8. Find two more solutions. How many solutions can we find?

Given equation:

To show:  are the solutions of the given equation.

Putting  in the equation, we get:

Now, putting  in the equation, we get:

Hence  are the solutions of .

Further solutions:
The equation can be rewritten as

Putting , we get:
Putting , we get:
Putting , we get: .....and so on
Two more  solutions are
Hence, we can find infinite solutions for the given equation.

#### Question 34:

For the equation 6x − 5y = 8, verify that
(i) (3, 2) is a solution
(ii) (2, 3) is not a solution

Given equation:

(i)  is a solution.
Putting  in the equation, we get:
.
Hence,  is a solution of the given equation.

(ii)  is not a solution.
Putting  in the equation, we get:
≠ $RHS$.
Hence,  is not a solution of the given equation.

#### Question 35:

If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.

Given equation: .
Also,  lies on the graph of the equation.
Putting  in the equation, we get:

⇒
⇒
⇒

#### Question 36:

Find two solutions for each of the following:
(i) 3x + 4y = 12
(ii) 3x + 5y = 0
(iii) 4y + 5 = 0

(i)
Or,

When , .
When , .
Hence , the two solutions are and .

(ii)
Or,
When , .

When , .
Hence, the two solution are $\left(5,-3\right)$ and .

(iii)
Or,
It is independent of $\text{x}$.
When  ,  .
When .
Hence, the two solutions are $\left(1,-\frac{5}{4}\right)$ and $\left(2,-\frac{5}{4}\right)$.

#### Question 37:

Study the graph given below. Choose the equation whose graph is given:
(i) y = x
(ii) y = 2x
(iii) y = 2x + 1
(iv) x + y = 0

(iii) y = 2x + 1

The two given points are .
These points must satisfy the equation in order to form the graph.
Let  be the equation of this graph.
At ,
So,
Now, at $B\left(-1,-1\right)$
So,
Hence, the given points satisfy this equation.
Hence,  is the equation of the graph.

#### Question 38:

Draw the graph of the equation 3x + 5y − 15 = 0 and show that x = 1, y = 2 is not a solution of the given equation.

Given equation:
Or,

When , then .

When , then .
Thus, we get the following table:

 $\text{x}$ 0 5 $y$ 3 0

Plot these points  on the graph paper. Join the points and extend them in both the directions.
Now, from the graph we can see that  does not lie on the line.
Moreover, we can check that  does not satisfy the equation as well.
≠ $RHS$.
Hence,  is not a solution of the given equation.

#### Question 39:

Draw the graph of the equation 3x + 2y = 12. At what points does the graph cut the x-axis and the y-axis?

Given equation:
or,

When , then .

When , then .

When , then .
Thus, we have the following table:

 $\text{x}$ 0 2 4 $y$ 6 3 0

Plot the points  and  on the graph paper.
Join the points and extend them in both the direction.
We see the graph cuts the $\text{x}$-axis at  and the $y$-axis at .

#### Question 40:

Draw the graph of the equation x − 2y = 6. Verify that each of the points P(2, −2), Q(4, −1) and R(−2, −4) lies on the straight line.

Given,
(or)

When , then .

When , then .

When , then  .

When , then .
Thus, we get the following table:

 $\text{x}$ 0 2 6 $y$ -  3 -2 0

Plot the points  on the graph paper. Join the points and extend them in both the directions.
From the graph, we can see that all the points lie on the straight line.

#### Question 41:

There are two scales of measuring temperature, namely, Fahrenheit (F) and Celsius (C).
The relation between the two scales is
$F=\frac{9}{5}C+32$
(i) Draw the graph of the given linear equation taking C along the x-axis and F along the y-axis.
Fill in the blanks given below:
(ii) 0°C = (......)°F
(iii) 95°F = (......)°C
(iv) 0°F = (......)°C
(v) Find the temperature which is numerically the same in both (F) and (C).

The given equation is .
On a graph paper, take $C$ along the $\text{x}$-axis and $F$ along the $y$-axis.
Along the $\text{x}$-axis, take 10 small divisions =
Along the $y$-axis, take 10 small divisions =
Putting  in the given equation, we get:
Putting  in the given equation, we get:
Thus, we get the following table:

 $\text{x}$ 0 5 $y$ 32 41

Plot the points  on the graph paper. Join these points and extend them in both the directions.
(ii) Putting  in the given equation, we get

(iii) Putting  in the given equation, we get:

⇒
⇒
∴

(iv) Putting  in the given equation, we get:

⇒

(v) Putting  in the given equation, we get:

⇒
⇒
∴

#### Question 42:

A taxi charges Rs 20 for the first km and Rs 12 per km for subsequent distance covered. Taking the distance covered as x km and total fare as Rs y, write a linear equation, depicting the relation in x and y.
Draw the graph between x and y.
From the graph, find the taxi charges for covering 16 km.

Taxi charges for the first km = Rs 20
Taxi charges for the subsequent distance covered = Rs 12 per km
Distance covered = x km  (given)
Total fare = Rs y (given)
Fare charged for the distance covered leaving the first kilometre  = Rs 12(x - 1)
​So, taxi charge = 20 + 12(x - 1)
But total fare = Rs y (given)
∴ y = 20 + 12x - 12 = 12x + 8
Hence, the required relation between x and y is y = 12x + 8
For,
For,
Plot these points  on the graph paper.
Join them and extend in both the directions.
Then,  gives .
So, for 16 km taxi charge = Rs 200.

#### Question 43:

If the work done by a body on applying a constant force is directly proportional to the distance travelled by the body, then express this in the form of an equation in two variables by taking the constant force as 4 units. From the graph, find the work done when the distance travelled is
(i) 2 units
(ii) 0 unit
(iii) 5 units

Let work done by the body = $W$
Constant force = 4 units
Let distance travelled by the body = $d$
According to the given condition, we have:
(∵ Work = Distance $×$ Force)
So,
when , then
when , then
Plot these points on the graph paper. Join the points and extend them in both the directions.
Then, we have:
(i) If , then  units
(ii) If , then  units
(iii) If , then  units

#### Question 1:

For the equation 5x + 8y = 50, if y = 10, the value of x is
(a) 6
(b) −6
(c) 12
(d) −12

(b) −6
Given equation:
If , then

⇒   .
⇒
Hence, .

#### Question 2:

The linear equation 2x + 5y = 16 has
(a) a unique solution
(b) two solutions
(c) no solution
(d) infinitely many solutions

(d) infinitely many solutions
The linear equation is   or,
When , then

When , then

When , then
Plotting these points  and  on the graph paper, we see that the equation has infinite solutions.

#### Question 3:

Express $\frac{2x}{3}+\frac{y}{6}-5=0$ in the form ax + by + c = 0.

Given,
Taking $LCM$ of the denominator, we get:

⇒   ,
which is the required form.

#### Question 4:

If 5y − 3x + 15 = 0, then express y in terms of x.

Given, .

which is the expression in $\text{x}$.

#### Question 5:

For what value of k does the point (k, −3) lie on the line 3x y = 6?

Given line:
Also $\left(k,-3\right)$ lies on the line.
Therefore, putting  in the equation of the line, we get:

Hence, .

#### Question 6:

If x = 3, y = −2 satisfy 2x − 3y = k, then find the value of k.

Given:  satisfy the equation . Then,

Hence, .

#### Question 7:

Find the points where the graph of the equation 3x + 4y = 12 cuts the x-axis and the y-axis.

Given equation:
or,
Now,
When , then

When , then

Thus, we have the following table:

 $\text{x}$ 0 4 $y$ 3 0

Thus, the graph of the equation cuts the $\text{x}$-axis at (4, 0) and the $y$-axis at (0,3).

#### Question 8:

The area of the triangle formed by the line x + 3y = 12 and the coordinate axes is
(a) 12 sq units
(b) 18 sq units
(c) 24 sq units
(d) 30 sq units

(c) 24 sq units
Given:   or,

When , then

When , then

When , then

When , then

When , then

Thus, we get the following table:
 $\text{x}$ 0 3 6 9 12 $y$ 4 3 2 1 0

Plot these points on the graph paper. Join them to get a right-angled triangle $AOE$, such that
unit and  unit.

Therefore,
sq units

#### Question 9:

Assertion: y = mx represents a line passing through the origin.
Reason: Any line parallel to the x-axis is y = k.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
We know that  represents a line passing through the origin.
So, Assertion (A) is true.
Moreover, we know that equation of a line parallel to the $\text{x}$-axis is .
So, Reason (R) is also true.
Hence both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

#### Question 10:

Assertion: x = 3 is a line parallel to the y-axis.
Reason: Any line parallel to the y-axis is y = k, where k R.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

(c) Assertion is true and Reason is false.
We know that  is a line parallel to $y$-axis.
Hence, Assertion (A) is true.
Any line parallel to the $y$-axis is of the form , where .
Hence Reason (R) is not true.
Hence, Assertion (A) is true and Reason (R) is false.

#### Question 11:

Match the following columns:

 Column I Column II (a) Any line parallel to the x-axis is (p) 3 (b) Any line parallel to the y-axis is (q) y = mx (c) Any line passing through the origin is (r) x = k (d) If the point (−2, 2) lies on the line ax + 4y = 2, then a = (s) y = k
(a) ......,
(b) ......,
(c) ......,
(d) ......,

(a) Any line parallel to the $\text{x}$-axis is .
(b) Any line parallel to the $y$-axis is .
(c) Any line passing through the origin is .
(d) The point (-2, 2) lies on the line .
Putting  in the equation, we get:

 Column I Column II (a) Any line parallel to the $\text{x}$-axis is (s) (b) Any line parallel to the -axis is (r) (c) Any line passing through the origin is (q) (d) If the point (-2, 2) lies on the line , then (p) 3

#### Question 12:

Give the geometrical representation of x = 3 as an equation in
(i) one variable
(ii) two variables

(i) Geometrical representation of  in one variable is .
(ii) Geometrical representation of  in two variable is .

#### Question 13:

For the line 2x + 3y = 6
(i) x-intercept = ......
(ii) y-intercept = ......

Line:
(i) Putting  in the equation , we get:

Hence, $\text{x}$-intercept = 3

(ii) Putting  in the equation, we get:

Hence, $y$-intercept = 2

#### Question 14:

Draw the graph of the line y = x and show that the point (2, 3) does not lie on it.

Equation of line : .
Then,

and so on.
Plot the points  .
Join the points and extend them in both the directions.

We see that  does not lie on the line .

#### Question 15:

Draw the graph of 2x − 3y = 4. From the graph, find whether x = −1, y = −2 is a solution or not.

Given:   or,

When , then

When , then

When , then

Thus, we get the following table:

 $\text{x}$ 2 5 -1 $y$ 0 2 -2

Plot the points  and  on the graph paper. Join them and extend them in both the directions.

Thus, we see that  is a solution of the equation.

#### Question 16:

The runs scored by two batsmen in a cricket match are 164. Write a linear equation in two variables x and y. Also write a solution of this equation.

Let one batsman score $\text{x}$ runs.
Then, the other batsman scores $y$ runs.
Given, total runs = 164
Then, according the given condition, we have:

Solution:
Putting  in the equation, we get:

Again, putting  in the equation, we get:

Hence,  are the two solutions of the equation.

#### Question 17:

Find whether the given statement is true or false:
(i) x = 2, y = 3 is a solution of the equation 5x − 3y = 1.
(ii) y = 2x + 5 is a straight line passing through the point (1, 5).
(iii) The area bounded by the line x + y = 6, the x-axis and the y-axis is 18 sq units.

(i) Given equation: .
To check wether  is the solution of the equation.
Putting  in the equation, we get:

Hence,  is the solution of the equation.
Hence, the statement is true.

(ii) Given equation:
When , then .
When , then .
When , then .
When , then .
When , then .

Plot these points on a graph paper and join them in both the directions.

From here, we see that for  we have ≠ 5. Moreover, from the graph we see that  does not lie on the line .
Hence, the statement is false.

(iii) Given : .
When , then .
When , then .
Plotting these points on the graph paper, we get the right-angled triangle$AOB$, such that  unit and  unit.

Hence,  sq unit.
Hence, the statement is true.

#### Question 18:

Two men start from points A and B respectively, 42 km apart. One walks from A to B at 4 km/hr and another walks from B to A at a certain uniform speed. They meet each other after 6 hours. Find the speed of the second man.

Given: Two men start from  respectively, which are 42 km apart.
Speed of man at $A$ = 4 km/hr.
Let the speed of man at $B$ = x km/hr.
They meet after 6 hours.
Then, relative speed of the two men =  km/hr

Therefore, speed of man at $B$ = Relative speed of the two - Speed of man at $A$ = 7 - 4 = 3 km/hr
Hence, speed of the second man = 3 km/hr.

#### Question 19:

The taxi fare in a city is such that Rs 50 is the fixed amount and Rs 16 per km is charged. Taking the distance covered as x km and total fare as Rs y, write a linear equation in x and y. What is the total fare for 20 km?

Given: Taxi fare in the city (fixed amount) = Rs 50
Taxi fare per km = Rs 16
Distance covered = x km
Total fare = Rs y
Therefore, fare charged per km = Rs 16x
Hence,
Total fare = y = 50 + 16x
Hence, the linear equation: y = 50 + 16x

Total fare for 20 km:
Putting x = 20 in the linear equation, we get:
y = 50 + 16$×$20 = 50 + 320 = Rs 370

#### Question 20:

Draw the graphs for the equations x + y = 6 and x y = 2 on the same graph paper and find the coordinates of the point where the two straight lines intersect.

The equations are  and .
Now, consider .
When , then .
When , then .
When , then .

Thus, we get the following table:

 $\text{x}$ 0 2 6 $y$ 6 4 0

Plot the points  and  on the graph paper. Then, join the points and extend them in both the directions.
Now, consider .
When , then .
When  then .
When , then .

Thus, we get the following table:
 $\text{x}$ 0 1 2 $y$ -2 -1 0

Plot these points on the graph paper. Then, join the points and extend them in both the directions.

From the graph, we can see that the two lines intersect at (4,2) .

View NCERT Solutions for all chapters of Class 9