# RS Aggarwal Solutions Class 10 Ex 3A

## RS Aggarwal Class 10 Ex 3A Chapter 3

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 – 5y + 4 = 0, 2x + y – 8 = 0

Solutions:

Given equations are 2x – 5y + 4 = 0 and 2x + y – 8 = 0

Graph of 2x – 5y + 4 = 0:

2x – 5y + 4 = 0 $\Rightarrow y = \frac{2x + 4}{5}$

Putting x = -2, we get y = 0

Putting x = 3, we get y = 2

Putting x = 8, we get y = 4

Hence, the table is,

 x -2 3 8 y 0 2 4

Plot the points A(-2, 0), B(3, 2) and C(8, 4) 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 – 5y + 4 = 0.

Graph of 2x + y – 8 = 0:

2x + y – 8 = 0 $\Rightarrow y = -2x + 8$

Putting x = 1, we get y = 6

Putting x = 3, we get y = 2

Putting x = 2, we get y = 4

Hence, the table is,

 x 1 3 2 y 6 2 4

Now, on the same graph paper as above plot the points P(1, 6) and Q(2, 4). The point B(3, 2) has already been plotted. Join PQ and QB. Thus, the line PB is the graph of 2x + y – 8 = 0.

The two graph lines intersect at B(3, 2). Therefore, x = 3, 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 + y + 1 = 0,

2x + y – 8 = 0

Solution:

Given equations are 3x + y + 1 = 0 and 2x + y – 8 = 0

Graph of 3x + y + 1 = 0:

3x + y + 1 = 0 $\Rightarrow y = -3 – 1$

Putting x = 0, we get y = -1

Putting x = -1, we get y = 2

Putting x = 1, we get y = -4

Hence, the table is,

 x 0 -1 1 y -1 2 -4

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

Thus, line AC is the graph of 3x + y + 1 = 0.

Graph of 2x + y – 8 = 0:

2x + y – 8 = 0 $\Rightarrow y = \frac{-2x + 8}{3}$

Putting x = -1, we get y = 2

Putting x = 2, we get y = 4

Putting x = -4, we get y = 0

Hence, the table is,

 x -1 2 -4 y 2 4 0

Now, on the same graph paper as above plot the points P(2, 4) and Q(-4, 0). The point B(-1, 2) has already been plotted. Join PB and BQ. Thus, the line PQ is the graph of 2x + y – 8 = 0.

The two graph lines intersect at B(-1, 2). Therefore, x = -1, y = 2 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.

3x – 2y + 2 = 0,

(3/2) x – y + 3 = 0

Solution:

Given equations are 3x – 2y + 2 = 0 and (3/2)x – y + 3 = 0

Graph of 3x – 2y + 2 = 0:

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

Putting x = 0, we get y = 1

Putting x = 2, we get y = 4

Putting x = 2, we get y = -2

Hence, the table is:

 x 0 2 -2 y 1 4 -2

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

Thus, line AC is the graph of 3x + y + 1 = 0.

Graph of (3/2)x – y + 3 = 0:

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

Putting x = 0, we get y = 3

Putting x = 2, we get y = 6

Putting x = -2, we get y = 0

Hence, the table is,

 x 0 2 -2 y 3 6 0

Now, on the same graph paper as above plot the points P(0, 3) and Q(2, 6) and R(-2, 0). Join PQ and PR to get QR. Thus, the line QR is the graph of (3/2)x – y + 3 = 0.

It is clear from the graph that the two lines are parallel and do not intersect even when produced.

Therefore, given equations are inconsistent has no solution.

The coordinates of the points where these, lines meet y-axis are A(0, 1) and B(0, 3) respectively.

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.

3x + y – 5 = 0,

2x – y – 5 = 0

Solution:

Given equations are 3x + y – 5 = 0 and 2x – y – 5 = 0

Graph of 3x + y – 5 = 0:

3x + y – 5 = 0 $\Rightarrow y = -3x + 5$

Putting x = 0, we get y = 5

Putting x = 1, we get y = 2

Putting x = 2, we get y = -1

Hence, the table is,

 x 0 1 2 y 5 2 -1

Plot the points A(0, 5), B(1, 2) and C(2, -1) 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 3x + y – 5 = 0.

Graph of 2x – y – 5 = 0:

2x – y – 5 = 0 $\Rightarrow y =2x – 5$

Putting x = 0, we get y = -5

Putting x = 1, we get y = -3

Putting x = 3, we get y = 1

Hence, the table is,

 x 0 1 3 y -5 -3 1

Now, on the same graph paper as above plot the points P(0, -5) and Q(1, -3) and R(3, 1). Join PQ and QR to get PR. Thus, the line PR is the graph of 2x – y – 5 = 0.

The lines (1) and (2) intersect y-axis at (0, 5) and (0, -5) respectively.

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.

3x + 5y = 15,

x – y = 5

Solution:

Given equations are 3x + 5y = 15 and x – y = 5

Graph of 3x + 5y = 15:

3x + 5y = 15 $\Rightarrow y = \frac{-3x + 15}{5}$

Putting x = 0, we get y = 3

Putting x = 5, we get y = 0

Putting x = -5, we get y = 6

Hence, the table is,

 x 0 5 -5 y 3 0 6

Plot the points A(0, 3), B(5, 0) and C(-5, 6) 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 + 5y = 15.

Graph of x – y = 5:

x – y = 5 $\Rightarrow y = x – 5$

Putting x = 0, we get y = -5

Putting x = 5, we get y = 0

Putting x = 2, we get y = -3

Hence, the table is,

 x 0 5 2 y -5 0 -3

Now, on the same graph paper as above plot the points P(0, -5) and Q(2, -3). Join PQ and QB to get PB. Thus, the line PB is the graph of x – y = 5.

It is clear from the graph that the given system of equations is consistent.

The lines 3x + 5y = 15 and x – y = 5 meet the y-axis at A(0, 3) and P(0, -5) respectively.

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.

x + 2y = 5,

2x – 3y = -4

Solution:

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

Graph of x + 2y = 5:

x + 2y = 5 $\Rightarrow y = \frac{5 – x}{2}$

Putting x = 1, we get y = 2

Putting x = 3, we get y = 1

Putting x = 5, we get y = 0

Hence, the table is,

 x 1 3 5 y 2 1 0

Plot the points A(1, 2), B(3, 1) and C(5, 0) 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 = 5.

Graph of 2x – 3y = -4:

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

Putting x = 1, we get y = 2

Putting x = -2, we get y = 0

Putting x = 4, we get y = 4

Hence, the table is,

 x 1 -2 4 y 2 0 4

Now, on the same graph paper as above plot the points P(4, 4) and Q(-2, 0). The point A(1, 2) has already been plotted. Join PA and QA to get PQ. the line PQ is the graph of the equation 2x – 3y = -4

The two graph lines intersect at point A(1, 2)

Therefore, x = 1, y = 2 is the solution of the given systems of equations. The region bounded by these lines and x-axis has been shaded.

On extending the graph lines on both sides, we find that these graph lines intersect x-axis at points Q(-2, 0) and C(5, 0)

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.

4x – 5y + 16 = 0, 2x + y – 6 =0

Solution:

Given equations are 4x – 5y + 16 = 0 and 2x + y – 6 =0

Graph of 4x – 5y + 16 = 0:

4x – 5y + 16 = 0 $\Rightarrow y = \frac{4x + 16}{5}$

Putting x = 1, we get y = 4

Putting x = -4, we get y = 0

Putting x = 6, we get y = 8

Hence, the table is,

 x 1 -4 6 y 4 0 8

Plot the points A(1, 4), B(-4, 0) and C(6, 8) on the graph paper. Join AB and AC to get the graph line BC. Extend it both ways.

Thus, line AC is the graph of 4x – 5y + 16 = 0.

Graph of 2x + y – 6 =0:

2x + y – 6 =0 $\Rightarrow y = -2x + 6$

Putting x = 1, we get y = 4

Putting x = 3, we get y = 0

Putting x = 2, we get y = 2

Hence, the table is,

 x 1 3 2 y 4 0 2

Now, on the same graph paper as above plot the points P(3, 0) and Q(2, 2). The point A(1, 4) has already been plotted. Join PQ and QA to get PA. the line PQ is the graph of the equation 2x + y – 6 =0. The two graph lines intersect at A(1, 4)

Therefore, x = 1, y = 4 is the solution of the given systems of equations. Clearly, the given equations are represented by the graph lines BC and PA respectively.

The vertices of $\Delta BAP$ formed by these lines and x-axis are B(-4, 0), A(1, 4) and P(3, 0).

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