Rs Aggarwal 2020 2021 Solutions for Class 9 Maths Chapter 4 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 Maths 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 2020 2021 Book of Class 9 Maths Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

#### Page No 145:

(i) 3x + 5= 7.5
This can be expressed in the form ax + by + c = 0 as $3x+5y+\left(-7.5\right)=0$.
(ii) $2x-\frac{y}{5}+6=0$
This can be expressed in the form ax + by + c = 0 as $2x+\left(-\frac{1}{5}\right)y+6=0$
(iii) 3y – 2x = 6
This can be expressed in the form ax + by + c = 0 as $2x+\left(-3y\right)+6=0$.
(iv) 4x = 5y
This can be expressed in the form ax + by + c = 0 as $4x-5y+0=0$
(v) $\frac{x}{5}-\frac{y}{6}=1$
This can be expressed in the form ax + by + c = 0 as $6x-5y=30$
(vi) $\sqrt{2}x+\sqrt{3}y=5$
This can be expressed in the form ax + by + c = 0 as $\sqrt{2}x+\sqrt{3}y+\left(-5\right)=0$

#### Page No 146:

(i) x = 6
In the form of ax + by + c = 0 we have $x+0y+\left(-6\right)=0$ where a = 1, b = 0 and $c=-6$.
(ii) 3x – y = x – 1
In the form of ax + by + c = 0 we have $2x+\left(-1y\right)+1=0$ where a = 2, b = $-1$ and c = 1
(iii) 2x + 9 = 0
In the form of ax + by + c = 0 we have $2x+\left(-1y\right)+1=0$ where a = 2, b = $-1$ and c = 1
(iv) 4y = 7
In the form of ax + by + c = 0 we have $0x+4y+\left(-7\right)=0$ where a = 0, b = 4 and c = $-7$
(v) x + y = 4
In the form of ax + by + c = 0 we have $1x+1y+\left(-4\right)=0$ where $a=1,b=1,c=-4$
(vi) $\frac{x}{2}-\frac{y}{3}=\frac{1}{6}+y$
In the form of ax + by + c = 0 we have $3x-2y=1+6y⇒3x-8y+\left(-1\right)=0\phantom{\rule{0ex}{0ex}}$ where $a=3,b=-8,c=-1$.

#### Page No 146:

The equation given is 5x – 4y = 20.
(i) (4, 0)
Putting the value in the given equation we have
$\mathrm{LHS}:5\left(4\right)-4\left(0\right)=20\phantom{\rule{0ex}{0ex}}\mathrm{RHS}:20\phantom{\rule{0ex}{0ex}}\mathrm{LHS}=\mathrm{RHS}$
Thus, (4, 0) is a solution of the given equation.
(ii) (0, 5)
Putting the value in the given equation we have
$\mathrm{LHS}:5\left(0\right)-4\left(5\right)=0-20=-20\phantom{\rule{0ex}{0ex}}\mathrm{RHS}:20\phantom{\rule{0ex}{0ex}}\mathrm{LHS}\ne \mathrm{RHS}$
Thus, (0, 5) is not a solution of the given equation.
(iii)
Putting the value in the given equation we have
$\mathrm{LHS}:5\left(-2\right)-4\left(\frac{5}{2}\right)=-10-10=-20\phantom{\rule{0ex}{0ex}}\mathrm{RHS}:20\phantom{\rule{0ex}{0ex}}\mathrm{LHS}\ne \mathrm{RHS}$
Thus, is not a solution of the given equation.
(iv) (0, –5)
Putting the value in the given equation we have
$\mathrm{LHS}:5\left(0\right)-4\left(-5\right)=0+20=20\phantom{\rule{0ex}{0ex}}\mathrm{RHS}:20\phantom{\rule{0ex}{0ex}}\mathrm{LHS}=\mathrm{RHS}$
Thus, (0, –5) is a solution of the given equation.
(v)
Putting the value in the given equation we have
$\mathrm{LHS}:5\left(2\right)-4\left(\frac{-5}{2}\right)=10+10=20\phantom{\rule{0ex}{0ex}}\mathrm{RHS}:20\phantom{\rule{0ex}{0ex}}\mathrm{LHS}=\mathrm{RHS}$
Thus,  is a solution of the given equation.

#### Page No 146:

(i) 2x – 3y = 6

 x 0 3 $-$3 $\frac{9}{2}$ 2 y $-$2 0 $-$4 1 $\frac{-2}{3}$

(ii) $\frac{2x}{5}+\frac{3y}{10}=3$
 x 0 $\frac{15}{2}$ 5 10 3 y 10 0 $\frac{10}{3}$ $\frac{-10}{3}$ 6

(iii) 3y = 4x

 x 3 $-$3 $-$6 6 0 y 4 $-$4 $-$8 8 0

#### Page No 146:

Given: 5x – 3y = k
Since x = 3 and = 4 is a solution of the given equation so, it should satisfy the equation.
$5\left(3\right)-3\left(4\right)=k\phantom{\rule{0ex}{0ex}}⇒15-12=k\phantom{\rule{0ex}{0ex}}⇒3=k$

#### Page No 146:

Given: 4– 3+ 1 = 0                                 .....(1)
x = 3k + 2 and y = 2k – 1
Putting these values in the equation (1) we get
$4\left(3k+2\right)-3\left(2k-1\right)+1=0\phantom{\rule{0ex}{0ex}}⇒12k+8-6k+3+1=0\phantom{\rule{0ex}{0ex}}⇒6k+12=0\phantom{\rule{0ex}{0ex}}⇒k+2=0\phantom{\rule{0ex}{0ex}}⇒k=-2$

#### Page No 146:

Let cost of a pencil to be ₹ x and that of a ballpoint to be ₹ y.
Cost of 5 pencils = 5x
Cost of 2 ballpoints = 2y
Cost of 5 pencils = cost of 2 ballpoints
$⇒5x=2y\phantom{\rule{0ex}{0ex}}⇒5x-2y=0$

#### Page No 161:

(i) The equation of given line is x = 4. This equation does not contain the term of y. So, the graph of this line is parallel to y-axis passing through the point (4, 0). (ii) The equation of given line is x + 4 = 0 or x = –4. This equation does not contain the term of y. So, the graph of this line is parallel to y-axis passing through the point (–4, 0). (iii) The equation of given line is y = 3. This equation does not contain the term of x. So, the graph of this line is parallel to x-axis passing through the point (0, 3). (iv) The equation of given line is y = –3. This equation does not contain the term of x. So, the graph of this line is parallel to x-axis passing through the point (0, –3). (v) The equation of given line is x = –2. This equation does not contain the term of y. So, the graph of this line is parallel to y-axis passing through the point (–2, 0). (vi) The equation of given line is x = 5. This equation does not contain the term of y. So, the graph of this line is parallel to y-axis passing through the point (5, 0). (vii) The equation of given line is y + 5 = 0 or y = –5. This equation does not contain the term of x. So, the graph of this line is parallel to x-axis passing through the point (0, –5). (viii) The equation of given line is y = 4. This equation does not contain the term of x. So, the graph of this line is parallel to x-axis passing through the point (0, 4). #### Page No 161:

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.
From the graph we can see that when x = 2, y = 6 Also, when x = -2, y = -6.

#### Page No 161:

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,

Similarly, from the graph we can see that when x = −5, y = 4.

#### Page No 161:

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: #### Page No 162:

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 $P\left(3,0\right)$.

#### Page No 162:

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).

#### Page No 162:

$3x-2y=4\phantom{\rule{0ex}{0ex}}⇒2y=3x-4\phantom{\rule{0ex}{0ex}}⇒y=\frac{3x-4}{2}$
When x = 0, $y=\frac{3×0-4}{2}=\frac{0-4}{2}=\frac{-4}{2}=-2$
When x = 2, $y=\frac{3×2-4}{2}=\frac{6-4}{2}=\frac{2}{2}=1$
When x = –2, $y=\frac{3×\left(-2\right)-4}{2}=\frac{-6-4}{2}=\frac{-10}{2}=-5$
Thus, the points on the line 3x – 2y = 4 are as given in the following table:

 x 0 2 –2 y –2 1 –5

Plotting the points (0, –2), (2, 1) and (–2–5) and drawing a line passing through these points, we obtain the graph of of the line 3x – 2y = 4.

$x+y-3=0\phantom{\rule{0ex}{0ex}}⇒y=-x+3$
When x = 0, $y=-0+3=3$
When x = 1, $y=-1+3=2$
When x = –1, $y=-\left(-1\right)+3=1+3=4$
Thus, the points on the line x + y – 3 = 0 are as given in the following table:
 x 0 1 –1 y 3 2 4

Plotting the points (0, 3), (1, 2) and (–14) and drawing a line passing through these points, we obtain the graph of of the line x + y – 3 = 0. It can be seen that the lines 3x – 2y = 4 and x + y – 3 = 0 intersect at the point (2, 1).

#### Page No 162:

$4x+3y=24\phantom{\rule{0ex}{0ex}}⇒3y=-4x+24\phantom{\rule{0ex}{0ex}}⇒y=\frac{-4x+24}{3}$
When x = 0, $y=\frac{-4×0+24}{3}=\frac{0+24}{3}=\frac{24}{3}=8$
When x = 3, $y=\frac{-4×3+24}{3}=\frac{-12+24}{3}=\frac{12}{3}=4$
When x = 6, $y=\frac{-4×6+24}{3}=\frac{-24+24}{3}=\frac{0}{3}=0$
Thus, the points on the line 4x + 3y = 24 are as given in the following table:

 x 0 3 6 y 8 4 0

Plotting the points (0, 8), (3, 4) and (60) and drawing a line passing through these points, we obtain the graph of of the line 4x + 3y = 24. (i) It can be seen that the line 4x + 3y = 24 intersects the x-axis at (6, 0) and y-axis at (0, 8).

(ii) The triangle formed by the line and the coordinate axes is a right triangle right angled at the origin.

∴ Area of the triangle = $\frac{1}{2}×6×8$ = 24 square units

#### Page No 162:

$2x+y=6\phantom{\rule{0ex}{0ex}}⇒y=-2x+6$
When x = 0, $y=-2×0+6=0+6=6$
When x = 1, $y=-2×1+6=-2+6=4$
When x = 2, $y=-2×2+6=-4+6=2$

Thus, the points on the line 2x + y = 6 are as given in the following table:

 x 0 1 2 y 6 4 2

Plotting the points (0, 6), (1, 4) and (22) and drawing a line passing through these points, we obtain the graph of of the line 2x + y = 6.

$2x-y+2=0\phantom{\rule{0ex}{0ex}}⇒y=2x+2$
When x = 0, $y=2×0+2=0+2=2$
When x = 1, $y=2×1+2=2+2=4$
When x = –1, $y=2×\left(-1\right)+2=-2+2=0$

Thus, the points on the line 2x – y + 2 = 0 are as given in the following table:
 x 0 1 –1 y 2 4 0

Plotting the points (0, 2), (1, 4) and (–10) and drawing a line passing through these points, we obtain the graph of of the line 2x – y + 2 = 0. The shaded region represents the area bounded by the lines 2x + y = 6, 2x – y + 2 = 0 and the x-axis. This represents a triangle.

It can be seen that the lines intersect at the point C(1, 4). Draw CD perpendicular from C on the x-axis.

Height = CD = 4 units

Base = AB = 4 units

∴ Area of the shaded region = Area of ∆ABC = $\frac{1}{2}×\mathrm{AB}×\mathrm{CD}=\frac{1}{2}×4×4$ = 8 square units

#### Page No 162:

$x-y=1\phantom{\rule{0ex}{0ex}}⇒y=x-1$
When x = 0, $y=0-1=-1$
When x = 1, $y=1-1=0$
When x = 2, $y=2-1=1$

Thus, the points on the line x – y = 1 are as given in the following table:

 x 0 1 2 y –1 0 1

Plotting the points (0, –1), (1, 0) and (21) and drawing a line passing through these points, we obtain the graph of of the line x – y = 1.

$2x+y=8\phantom{\rule{0ex}{0ex}}⇒y=-2x+8$
When x = 1, $y=-2×1+8=-2+8=6$
When x = 2, $y=-2×2+8=-4+8=4$
When x = 3, $y=-2×3+8=-6+8=2$

Thus, the points on the line  2x + y = 8 are as given in the following table:
 x 1 2 3 y 6 4 2

Plotting the points (1, 6), (2, 4) and (32) and drawing a line passing through these points, we obtain the graph of of the line  2x + y = 8. The shaded region represents the area bounded by the lines x – y = 1, 2x + y = 8 and the y-axis. This represents a triangle.

It can be seen that the lines intersect at the point C(3, 2). Draw CD perpendicular from C on the y-axis.

Height = CD = 3 units

Base = AB = 9 units

∴ Area of the shaded region = Area of ∆ABC = $\frac{1}{2}×\mathrm{AB}×\mathrm{CD}=\frac{1}{2}×9×3$ = $\frac{27}{2}$ square units

#### Page No 162:

$x+y=6\phantom{\rule{0ex}{0ex}}⇒y=-x+6$
When x = 0, $y=-0+6=6$
When x = 1, $y=-1+6=5$
When x = 3, $y=-3+6=3$

Thus, the points on the line x + y = 6 are as given in the following table:

 x 0 1 3 y 6 5 3

Plotting the points (0, 6), (1, 5) and (33) and drawing a line passing through these points, we obtain the graph of of the line x + y = 6.

$x-y=2\phantom{\rule{0ex}{0ex}}⇒y=x-2$
When x = 0, $y=0-2=-2$
When x = 2, $y=2-2=0$
When x = –1, $y=-1-2=-3$

Thus, the points on the line x – y = 2 are as given in the following table:
 x 0 2 –1 y –2 0 –3

Plotting the points (0, –2), (2, 0) and (–1–3) and drawing a line passing through these points, we obtain the graph of of the line x – y = 2. It can be seen that the lines x + y = 6 and x – y = 2 intersect at the point (4, 2).

#### Page No 162:

Let the contribution of A and B be ₹ and ₹ y, respectively.

Total contribution of A and B = ₹ + ₹ = ₹ (y)

It is given that the total contribution of A and B is ₹ 100.

∴ y = 100

This is the linear equation satisfying the the given data.

= 100

⇒ y = 100 – x

When x = 10, y = 100 – 10 = 90

When x = 40, y = 100 – 40 = 60

When x = 60, y = 100 – 60 = 40

Thus, the points on the line = 100  are as given in the following table:

 x 10 40 60 y 90 60 40

Plotting the points (10, 90), (40, 60) and (6040) and drawing a line passing through these points, we obtain the graph of of the line = 100. #### Page No 162:

Since, the y-coordinate of any point on x-axis is always 0.

So, the equation of the x-axis is y = 0.

Hence, the correct option is (b).

#### Page No 162:

Since, the x-coordinate of any point on y-axis is always 0.

So, the equation of the y-axis is x = 0.

Hence, the correct option is (a).

#### Page No 163:

(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 .

#### Page No 163:

(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 .

#### Page No 163:

(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. #### Page No 163:

Since, every point on the line represented by the equation 2x + 5= 7 is its solution.

Therefore, there are infinite solutions of the equation the equation 2x + 5= 7 in which the values of x and y are rational numbers, positive real numbers or real numbers.

But, as 2 + 5 = 7, i.e. x = 1 and y = 1 are the only pair of natural numbers that are the solution of the equation the equation 2x + 5= 7.

So, the equation 2x + 5= 7 has a unique solution, if x and are both are natural numbers.

Hence, the correct option is (a).

#### Page No 163:

As, the graph of y = 5 is a line parallel to x-axis i.e. y = 0.

$⇒$ The line represented by the equation y = 5 is parallel to x-axis and intersects y-axis at y = 5.

So, the graph of y = 5 is a line parallel to the x-axis at a distance of 5 units from the origin making an intercept 5 on the y-axis.

Hence, the correct answers are options (b) and (c).
Disclaimer: In this question, there are two correct answers.

#### Page No 163:

As, the graph of x = 4 is a line parallel to y-axis i.e. x = 0.

$⇒$ The line represented by the equation x = 4 is parallel to y-axis and intersects x-axis at x = 4.

So, the graph of x = 4 is parallel to y-axis at a distance of 4 units from the origin making an intercept 4 on the x-axis.

Hence, the correct options are (a) and (d).

Disclaimer: In this question, there are two correct answers.

#### Page No 163:

As, the graph of x + 3 = 0 or x = $-$3 is a line parallel to y-axis i.e. x = 0.

$⇒$ The line represented by the equation x = $-$3 is parallel to y-axis and intersects x-axis at x = $-$3.

So, the graph of x + 3 = 0 is a line parallel to the y-axis at a distance of 3 units to the left of y-axis making an intercept $-$3 on the x-axis.

Hence, the correct options are (a) and (c).

Disclaimer: In this question, there are two correct answers.

#### Page No 163:

As, the graph of y + 2 = 0 or y = −2 is a line parallel to x-axis i.e. y = 0.

⇒ The line represented by the equation y = −2 is parallel to x-axis and intersects y-axis at y = −2.

So, the graph of y + 2 = 0 is a line parallel to the x-axis at a distance of 2 units below the x-axis making an intercept −2 on the y-axis.

Hence, the correct options are (b) and (c).

Disclaimer: In this question, there are two correct answers.

#### Page No 163:

If he graph of the linear equation 2x + 3= 6 meets the y-axis, then x = 0.

Substituting the value of x = 0 in equation 2x + 3= 6, we get

$2\left(0\right)+3y=6\phantom{\rule{0ex}{0ex}}⇒3y=6\phantom{\rule{0ex}{0ex}}⇒y=\frac{6}{3}\phantom{\rule{0ex}{0ex}}⇒y=2$

So, the point of meeting is (0, 2).

Hence, the correct answer is option (c).

#### Page No 164:

If he graph of the linear equation 2x + 5= 10 meets the x-axis, then y = 0.

Substituting the value of y = 0 in equation 2x + 5= 10, we get

$2x+5\left(0\right)=10\phantom{\rule{0ex}{0ex}}⇒2x=10\phantom{\rule{0ex}{0ex}}⇒x=\frac{10}{2}\phantom{\rule{0ex}{0ex}}⇒x=5$

So, the point of meeting is (5, 0).

Hence, the correct option is (c).

#### Page No 164:

(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.

#### Page No 164:

Since, the graph of the line y = 3 is parallel to x-axis at a distance of 3 units from the x-axis.

Or, the y-coordinate of every point on the line is always equal to 3.

So, the graph of the line y = 3 passes through the point (2, 3).

Hence, the correct option is (c).

#### Page No 164:

(d) (−3, 2)

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

#### Page No 164:

(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).

#### Page No 164:

(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 .

#### Page No 164:

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

#### Page No 164:

(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.

#### Page No 164:

Since, (2, 0) is a solution of the linear equation 2x + 3y = k.

Substituting the values x = 2 and y = 0 in the equation 2x + 3y = k, we get

Hence, the correct option is (d).

#### Page No 164:

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

#### Page No 164:

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

#### Page No 164:

Substituting the values x = 5, y = 2 in

(a) x + 2y = 7, we get

(b) 5x + 2y = 7, we get

(c) x + y = 7, we get

(d) 5x + y = 7, we get

Hence, the correct option is (c).