# RS Aggarwal Solutions Class 10 Linear Equations In Two Variables

A linear equation in two variables, contains two different coefficients along with two variables. Linear equations are primarily represented in two forms which are basically Cartesian coordinates, Slope-Intercept form, Intercept form, Two-point form, Matrix form and Parametric form. The linear equation which is usually written as y= f(x), can be additivity and homogeneity. The general format of a linear equation in two variables can be represented as:

ax+ by + c= 0

Learn more about RS Aggarwal Class 10 Solutions Chapter 3 Linear equation in two variables, available below:

Question 1: On a graph paper, draw a horizontal line X’OX and a vertical line YOY’ as the x-axis and the y-axis respectively.

2x + 3y = 2, x – 2y = 8

Solution:

Given equations are 2x + 3y = 2 and x – 2y = 8

Graph of 2x + 3y = 2:

$y = \frac{2(1 – x)}{3}$

Putting x = 1, we get y = 0

Putting x = -2, we get y = 2

Putting x = 4, we get y = -2

Hence, the table is:

 x 1 -2 4 y 0 2 -2

Plot the points A (1, 0), B (-2, 2) and C (4, -2) on the graph paper. Join AB and AC to get the graph line BC. Extend it both ways.

Thus, line BC is the graph of 2x + 3y = 2.

Graph of x – 2y = 8:

$y = \frac{x – 8}{2}$

Putting x = 2, we get y = -3

Putting x = 4, we get y = -2

Putting x = 0, we get y = -4

Hence, the table is:

 x 2 4 0 y -3 -2 -4

Now, on the same graph paper as above plot the points P (0, -4) and Q (2, -3). The point C (4, -2) has already been plotted. Join QC and extend it. Thus, the line PC is the graph of x – 2y = 8.

The two graph lines intersect at C (4, -2). Therefore, x = 4, y = -2 is the solution of given system of equations.

Question 2: On a graph paper, draw a horizontal line X’OX and a vertical line YoY’ as the x-axis and the y-axis respectively.

3x + 2y = 4 and 2x – 3y = 7

Solution:

Given equations are 3x + 2y = 4 and 2x – 3y = 7

Graph of 3x + 2y = 4:

$y = \frac{4 – 3x}{2}$

Putting x = 0, we get y = 2

Putting x = 2, we get y = -1

Putting x = -2, we get y = 5

Hence, the table is:

 x 0 2 -2 y 2 -1 5

Plot the points A (0, 2), B (2, -1) and C (-2, 5) on the graph paper. Join AB and AC to get the graph line BC. Extend it both ways.

Thus, line BC is the graph of 3x + 2y = 4.

Graph of 2x – 3y = 7:

$y = \frac{x – 8}{2}$

Putting x = 2, we get y = -1

Putting x = -1, we get y = -3

Putting x = 5, we get y = 1

Hence, the table is:

 x 2 -1 5 y -1 -3 1

Now, on the same graph paper as above plot the points P (-1, -3) and Q (5, 1). The point B (2, -1) has already been plotted. Join QB and extend it. Thus, the line PQ is the graph of 2x – 3y = 7.

The two graph lines intersect at B (2,-1). Therefore, x = 2, y = -1 is the solution of given system of equations.

Question 3: On a graph paper, draw a horizontal line X’OX and a vertical line YoY’ as the x-axis and the y-axis respectively.

x – y + 1 = 0,

3x + 2y – 12 = 0

Solution:

Given equations are x – y + 1 = 0 and 3x + 2y – 12 = 0

Graph of x – y + 1 = 0:

x – y + 1 = 0 $\Rightarrow$ y = x + 1 ……..(1)

Putting x = 0, we get y = 1

Putting x = -1, we get y = 0

Putting x = 2, we get y = 3

Hence, the table is:

 x 0 -1 2 y 1 0 3

Plot the points A (0, 1), B (-1, 0) and C (2, 3) on the graph paper. Join AB and AC to get the graph line BC. Extend it both ways.

Thus, line BC is the graph of x – y + 1 = 0.

Graph of 3x + 2y – 12 = 0:

$y = \frac{12 – 3x}{2}$

Putting x = 0, we get y = 6

Putting x = 2, we get y = 3

Putting x = 4, we get y = 0

Hence, the table is:

 x 0 2 4 y 6 3 0

Now, on the same graph paper as above plot the points P (0, 6) and Q (4, 0). The point C (2, 3) has already been plotted. Join QB and extend it. Thus, the line PQ is the graph of 3x + 2y – 12 = 0.

The two graph lines intersect at (2, 3). Therefore, x = 2, y = 3 is the solution of given system of equations.

Question 4: On a graph paper, draw a horizontal line X’OX and a vertical line YoY’ as the x-axis and the y-axis respectively.

2x + 3y = 4, 3x – y = -5

Solution:

Given equations are 2x + 3y = 4 and 3x – y = -5

Graph of 2x + 3y = 4:

2x + 3y = 4 $\Rightarrow y = \frac{4 – 2x}{3}$

Putting x = -1, we get y = 2

Putting x = 2, we get y = 0

Putting x = 5, we get y = -2

Hence, the table is:

 x -1 2 5 y 2 0 -2

Plot the points A (-1, 2), B (2, 0) and C (5, -2) on the graph paper. Join AB and BC to get the graph line AC. Extend it both ways.

Thus, line BC is the graph of 2x + 3y = 4.

Graph of 3x – y = -5:

y = 3x + 5

Putting x = -1, we get y = -2

Putting x = 0, we get y = 5

Putting x = -2, we get y = -1

Hence, the table is:

 x -1 0 -2 y 2 5 -1

Now, on the same graph paper as above plot the points P(0, 5) and Q(-2, -1). The point A (-1, 2) has already been plotted. Join PA and amd QA to get line PQ. Thus, the line PQ is the graph of 3x – y = -5.

The two graph lines intersect at A (-1, 2). Therefore, x = -1, y = 2 is the solution of given system of equations.

Question 5: On a graph paper, draw a horizontal line X’OX and a vertical line YoY’ as the x-axis and the y-axis respectively.

2x – 3y = 1, 3x – 4y = 1

Solution:

Given equations are 2x – 3y = 1 and 3x – 4y = 1

Graph of 2x – 3y = 1:

2x – 3y = 1 $\Rightarrow y = \frac{2x – 1}{3}$

Putting x = -1, we get y = -1

Putting x = 2, we get y = 1

Putting x = 5, we get y = 3

Hence, the table is:

 x -1 2 5 y -1 1 3

Plot the points A (-1, -1), B (2, 1) and C (5, 3) on the graph paper. Join AB and BC to get the graph line AC. Extend it both ways.

Thus, line AC is the graph of 2x – 3y = 1.

Graph of 3x – 4y = 1:

3x – 4y = 1 $\Rightarrow y = \frac{3x – 1}{4}$

Putting x = -1, we get y = -1

Putting x = 3, we get y = 2

Putting x = 2, we get y = -4

Hence, the table is:

 x -1 3 -5 y -1 2 -4

Now, on the same graph paper as above plot the points P (3, 2) and Q(-5, -4). The point A (-1, -1) has already been plotted. Join PA and QA. Thus, the line PQ is the graph of 3x – 4y = 1.

The two graph lines intersect at A (-1, -1). Therefore, x = -1, y = -1 is the solution of given system of equations.

Question 6: On a graph paper, draw a horizontal line X’OX and a vertical line YoY’ as the x-axis and the y-axis respectively.

4x + 3y = 5, 2y – x = 7

Solution:

Given equations are 4x + 3y = 5 and 2y – x = 7

Graph of 4x + 3y = 5:

4x + 3y = 5 $\Rightarrow y = \frac{5 – 4x}{3}$

Putting x = -1, we get y = 3

Putting x = 2, we get y = -1

Putting x = 5, we get y = -5

Hence, the table is:

 x -1 2 5 y 3 -1 -5

Plot the points A (-1, 3), B (2, -1) and C (5, -5) on the graph paper. Join AB and BC to get the graph line AC. Extend it both ways.

Thus, line AC is the graph of 4x + 3y = 5.

Graph of 2y – x = 7:

2y – x = 7$\Rightarrow y = \frac{7 + x}{2}$

Putting x = -1, we get y = 3

Putting x = 3, we get y = 5

Putting x = -3, we get y = 2

Hence, the table is:

 x -1 3 -3 y 3 5 2

Now, on the same graph paper as above plot the points P (3, 5) and Q (-3, 2). The point A (-1, 3) has already been plotted. Join PA and PQ. Thus, the line PQ is the graph of 2y – x = 7.

The two graph lines intersect at A(-1, -3). Therefore, x = -1, y = 3 is the solution of given system of equations.

Question 7: On a graph paper, draw a horizontal line X’OX and a vertical line YoY’ as the x-axis and the y-axis respectively.

x + 2y + 2 = 0, 3x + 2y + 2 = 0

Solution:

Given equations are x + 2y + 2 = 0 and 3x + 2y + 2 = 0

Graph of x + 2y + 2 = 0:

x + 2y + 2 = 0 $\Rightarrow y = \frac{-x -2}{2}$

Putting x = -2, we get y = 0

Putting x = 0, we get y = -1

Putting x = 2, we get y = -2

Hence, the table is:

 x -2 0 2 y 0 -1 -2

Plot the points A (-2, 0), B (0, -1) and C (2, -2) on the graph paper. Join AB and BC to get the graph line AC. Extend it both ways.

Thus, line AC is the graph of x + 2y + 2 = 0.

Graph of 3x + 2y + 2 = 0:

3x + 2y + 2 = 0 $\Rightarrow y = \frac{-3x + 2}{2}$

Putting x = 0, we get y = 1

Putting x = 2, we get y = -2

Putting x = 4, we get y = -5

Hence, the table is:

 x 0 2 4 y 1 -2 -5

Now, on the same graph paper as above plot the points P (0, 1) and Q (4, -5). The point C (2, -2) has already been plotted. Join PC and QC. Thus, the line PQ is the graph of 3x + 2y + 2 = 0.

The two graph lines intersect at C(2, -2). Therefore, x = 2, y = -2 is the solution of given system of equations.

Question 8: On a graph paper, draw a horizontal line X’OX and a vertical line YoY’ as the x-axis and the y-axis respectively.

2x + 3y = 8, x – 2y + 3 = 0

Solution:

Given equations are 2x + 3y = 8 and x – 2y + 3 = 0

Graph of 2x + 3y = 8:

2x + 3y = 8 $\Rightarrow y = \frac{8 – 2x}{3}$

Putting x = 1, we get y = 2

Putting x = -5, we get y = 6

Putting x = 7, we get y = -2

Hence, the table is:

 x 1 -5 7 y 2 6 -2

Plot the points A (1, 2), B (-5, 6) and C (7, -2) on the graph paper. Join AB and BC to get the graph line BC. Extend it both ways.

Thus, line AC is the graph of 2x + 3y = 8.

Graph of x – 2y + 3 = 0:

x – 2y + 3 = 0 $\Rightarrow y = \frac{x + 3}{2}$

Putting x = 1, we get y = 2

Putting x = 3, we get y = 3

Putting x = -3, we get y = 0

Hence, the table is:

 x 1 3 -3 y 2 3 0

Now, on the same graph paper as above plot the points P (3, 3) and Q (-3, 0). The point A (1, 2) has already been plotted. Join PA and QA. Thus, the line PQ is the graph of x – 2y + 3 = 0.

The two graph lines intersect at A (1, 2). Therefore, x = 1, y = 2 is the solution of given system of equations.

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