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# RS Aggarwal Solutions: Exercise 3A - Linear Equations in two variables Notes | EduRev

## Class 10 : RS Aggarwal Solutions: Exercise 3A - Linear Equations in two variables Notes | EduRev

``` Page 1

Exercise – 3A

1. Solve the system of equations graphically:
2x + 3y = 2,
x – 2y = 8
Sol:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' representing the x-
axis and y-axis, respectively.
Graph of 2x + 3y = 2
2x + 3y = 2
?3y = (2 – 2x)
?3y = 2(1 – x)
?y =
?? (?? -?? )
??           …(i)
Putting x = 1, we get y = 0
Putting x = -2, we get y = 2
Putting x = 4, we get y = -2
Thus, we have the following table for the equation 2x + 3y = 2
x 1 -2 4
y 0 2 -2
Now, 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 on both ways.
Thus, the line BC is the graph of 2x + 3y = 2.

Graph of x - 2y = 8
x – 2y = 8
? 2y = (x – 8)
? y =
?? -?? ??            …(ii)
Putting x = 2, we get y = -3
Putting x = 4, we get y = -2
Putting x = 0, we get y = -4
Thus, we have the following table for the equation x – 2y = 8.
x 2 4 0
y -3 -2 -
4
Now, plot the points P(0, -4) and Q(2, -3). The point C(4, -2) has already been plotted. Join
PQ and QC and extend it on both ways.
Thus, line PC is the graph of x – 2y = 8.

Page 2

Exercise – 3A

1. Solve the system of equations graphically:
2x + 3y = 2,
x – 2y = 8
Sol:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' representing the x-
axis and y-axis, respectively.
Graph of 2x + 3y = 2
2x + 3y = 2
?3y = (2 – 2x)
?3y = 2(1 – x)
?y =
?? (?? -?? )
??           …(i)
Putting x = 1, we get y = 0
Putting x = -2, we get y = 2
Putting x = 4, we get y = -2
Thus, we have the following table for the equation 2x + 3y = 2
x 1 -2 4
y 0 2 -2
Now, 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 on both ways.
Thus, the line BC is the graph of 2x + 3y = 2.

Graph of x - 2y = 8
x – 2y = 8
? 2y = (x – 8)
? y =
?? -?? ??            …(ii)
Putting x = 2, we get y = -3
Putting x = 4, we get y = -2
Putting x = 0, we get y = -4
Thus, we have the following table for the equation x – 2y = 8.
x 2 4 0
y -3 -2 -
4
Now, plot the points P(0, -4) and Q(2, -3). The point C(4, -2) has already been plotted. Join
PQ and QC and extend it on both ways.
Thus, line PC is the graph of x – 2y = 8.

The two graph lines intersect at C(4, -2).
? x = 4 and y = -2 are the solutions of the given system of equations.

2. Solve the system of equations graphically:
3x + 2y = 4,
2x – 3y = 7
Sol:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' representing the x-
axis and y-axis, respectively.
Graph of 3x + 2y = 4
3x + 2y = 4
?2y = (4 – 3x)
?y =
?? - ???? )
??           …(i)
Putting x = 0, we get y = 2
Putting x = 2, we get y = -1
Putting x = -2, we get y = 5
Thus, we have the following table for the equation 3x + 2y = 4
x 0 2 -2
y 2 -1 5
Now, 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 on both ways.
Thus, BC is the graph of 3x + 2y = 4.

Graph of 2x - 3y = 7
2x – 3y = 7
? 3y = (2x – 7)
? y =
???? - ?? ??            …(ii)
Putting x = 2, we get y = -1
Putting x = -1, we get y = -3
Putting x = 5, we get y = 1
Page 3

Exercise – 3A

1. Solve the system of equations graphically:
2x + 3y = 2,
x – 2y = 8
Sol:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' representing the x-
axis and y-axis, respectively.
Graph of 2x + 3y = 2
2x + 3y = 2
?3y = (2 – 2x)
?3y = 2(1 – x)
?y =
?? (?? -?? )
??           …(i)
Putting x = 1, we get y = 0
Putting x = -2, we get y = 2
Putting x = 4, we get y = -2
Thus, we have the following table for the equation 2x + 3y = 2
x 1 -2 4
y 0 2 -2
Now, 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 on both ways.
Thus, the line BC is the graph of 2x + 3y = 2.

Graph of x - 2y = 8
x – 2y = 8
? 2y = (x – 8)
? y =
?? -?? ??            …(ii)
Putting x = 2, we get y = -3
Putting x = 4, we get y = -2
Putting x = 0, we get y = -4
Thus, we have the following table for the equation x – 2y = 8.
x 2 4 0
y -3 -2 -
4
Now, plot the points P(0, -4) and Q(2, -3). The point C(4, -2) has already been plotted. Join
PQ and QC and extend it on both ways.
Thus, line PC is the graph of x – 2y = 8.

The two graph lines intersect at C(4, -2).
? x = 4 and y = -2 are the solutions of the given system of equations.

2. Solve the system of equations graphically:
3x + 2y = 4,
2x – 3y = 7
Sol:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' representing the x-
axis and y-axis, respectively.
Graph of 3x + 2y = 4
3x + 2y = 4
?2y = (4 – 3x)
?y =
?? - ???? )
??           …(i)
Putting x = 0, we get y = 2
Putting x = 2, we get y = -1
Putting x = -2, we get y = 5
Thus, we have the following table for the equation 3x + 2y = 4
x 0 2 -2
y 2 -1 5
Now, 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 on both ways.
Thus, BC is the graph of 3x + 2y = 4.

Graph of 2x - 3y = 7
2x – 3y = 7
? 3y = (2x – 7)
? y =
???? - ?? ??            …(ii)
Putting x = 2, we get y = -1
Putting x = -1, we get y = -3
Putting x = 5, we get y = 1

Thus, we have the following table for the equation 2x – 3y = 7.

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

The two graph lines intersect at B(2, -1).
?x = 2 and y = -1 are the solutions of the given system of equations.

3. Solve the system of equations graphically:
2x + 3y = 8,
x – 2y + 3 = 0
Sol:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and
y-axis, respectively.
Graph of 2x + 3y = 8
2x + 3y = 8
?3y = (8 – 2x)
?y =
?? - ????
??           …(i)
Putting x = 1, we get y = 2.
Putting x = -5, we get y = 6.
Putting x = 7, we get y = -2.
Thus, we have the following table for the equation 2x + 3y = 8.
x 1 -5 7
y 2 6 -2
Now, plot the points A(1, 2), B(5, -6) and C(7, -2) on the graph paper.
Join AB and AC to get the graph line BC. Extend it on both ways.
Thus, BC is the graph of 2x + 3y = 8.
Page 4

Exercise – 3A

1. Solve the system of equations graphically:
2x + 3y = 2,
x – 2y = 8
Sol:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' representing the x-
axis and y-axis, respectively.
Graph of 2x + 3y = 2
2x + 3y = 2
?3y = (2 – 2x)
?3y = 2(1 – x)
?y =
?? (?? -?? )
??           …(i)
Putting x = 1, we get y = 0
Putting x = -2, we get y = 2
Putting x = 4, we get y = -2
Thus, we have the following table for the equation 2x + 3y = 2
x 1 -2 4
y 0 2 -2
Now, 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 on both ways.
Thus, the line BC is the graph of 2x + 3y = 2.

Graph of x - 2y = 8
x – 2y = 8
? 2y = (x – 8)
? y =
?? -?? ??            …(ii)
Putting x = 2, we get y = -3
Putting x = 4, we get y = -2
Putting x = 0, we get y = -4
Thus, we have the following table for the equation x – 2y = 8.
x 2 4 0
y -3 -2 -
4
Now, plot the points P(0, -4) and Q(2, -3). The point C(4, -2) has already been plotted. Join
PQ and QC and extend it on both ways.
Thus, line PC is the graph of x – 2y = 8.

The two graph lines intersect at C(4, -2).
? x = 4 and y = -2 are the solutions of the given system of equations.

2. Solve the system of equations graphically:
3x + 2y = 4,
2x – 3y = 7
Sol:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' representing the x-
axis and y-axis, respectively.
Graph of 3x + 2y = 4
3x + 2y = 4
?2y = (4 – 3x)
?y =
?? - ???? )
??           …(i)
Putting x = 0, we get y = 2
Putting x = 2, we get y = -1
Putting x = -2, we get y = 5
Thus, we have the following table for the equation 3x + 2y = 4
x 0 2 -2
y 2 -1 5
Now, 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 on both ways.
Thus, BC is the graph of 3x + 2y = 4.

Graph of 2x - 3y = 7
2x – 3y = 7
? 3y = (2x – 7)
? y =
???? - ?? ??            …(ii)
Putting x = 2, we get y = -1
Putting x = -1, we get y = -3
Putting x = 5, we get y = 1

Thus, we have the following table for the equation 2x – 3y = 7.

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

The two graph lines intersect at B(2, -1).
?x = 2 and y = -1 are the solutions of the given system of equations.

3. Solve the system of equations graphically:
2x + 3y = 8,
x – 2y + 3 = 0
Sol:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and
y-axis, respectively.
Graph of 2x + 3y = 8
2x + 3y = 8
?3y = (8 – 2x)
?y =
?? - ????
??           …(i)
Putting x = 1, we get y = 2.
Putting x = -5, we get y = 6.
Putting x = 7, we get y = -2.
Thus, we have the following table for the equation 2x + 3y = 8.
x 1 -5 7
y 2 6 -2
Now, plot the points A(1, 2), B(5, -6) and C(7, -2) on the graph paper.
Join AB and AC to get the graph line BC. Extend it on both ways.
Thus, BC is the graph of 2x + 3y = 8.

Graph of x - 2y + 3 = 0
x – 2y + 3 = 0
? 2y = (x + 3)
? y =
?? + ?? ??            …(ii)
Putting x = 1, we get y = 2.
Putting x = 3, we get y = 3.
Putting x = -3, we get y = 0.
Thus, we have the following table for the equation x – 2y + 3 = 0.
x 1 3 -3
y 2 3 0
Now, plot the points P (3, 3) and Q (-3, 0). The point A (1, 2) has already been plotted. Join
AP and QA and extend it on both ways.
Thus, PQ is the graph of x – 2y + 3 = 0.

The two graph lines intersect at A (1, 2).
? x = 1 and y = 2.

4. Solve the system of equations graphically:
2x - 5y + 4 = 0,
2x + y - 8 = 0
Sol:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and
y-axis, respectively.
Graph of 2x - 5y + 4 = 0
2x – 5y + 4 = 0
?5y = (2x + 4)
?y =
???? + ?? ??           …(i)
Page 5

Exercise – 3A

1. Solve the system of equations graphically:
2x + 3y = 2,
x – 2y = 8
Sol:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' representing the x-
axis and y-axis, respectively.
Graph of 2x + 3y = 2
2x + 3y = 2
?3y = (2 – 2x)
?3y = 2(1 – x)
?y =
?? (?? -?? )
??           …(i)
Putting x = 1, we get y = 0
Putting x = -2, we get y = 2
Putting x = 4, we get y = -2
Thus, we have the following table for the equation 2x + 3y = 2
x 1 -2 4
y 0 2 -2
Now, 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 on both ways.
Thus, the line BC is the graph of 2x + 3y = 2.

Graph of x - 2y = 8
x – 2y = 8
? 2y = (x – 8)
? y =
?? -?? ??            …(ii)
Putting x = 2, we get y = -3
Putting x = 4, we get y = -2
Putting x = 0, we get y = -4
Thus, we have the following table for the equation x – 2y = 8.
x 2 4 0
y -3 -2 -
4
Now, plot the points P(0, -4) and Q(2, -3). The point C(4, -2) has already been plotted. Join
PQ and QC and extend it on both ways.
Thus, line PC is the graph of x – 2y = 8.

The two graph lines intersect at C(4, -2).
? x = 4 and y = -2 are the solutions of the given system of equations.

2. Solve the system of equations graphically:
3x + 2y = 4,
2x – 3y = 7
Sol:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' representing the x-
axis and y-axis, respectively.
Graph of 3x + 2y = 4
3x + 2y = 4
?2y = (4 – 3x)
?y =
?? - ???? )
??           …(i)
Putting x = 0, we get y = 2
Putting x = 2, we get y = -1
Putting x = -2, we get y = 5
Thus, we have the following table for the equation 3x + 2y = 4
x 0 2 -2
y 2 -1 5
Now, 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 on both ways.
Thus, BC is the graph of 3x + 2y = 4.

Graph of 2x - 3y = 7
2x – 3y = 7
? 3y = (2x – 7)
? y =
???? - ?? ??            …(ii)
Putting x = 2, we get y = -1
Putting x = -1, we get y = -3
Putting x = 5, we get y = 1

Thus, we have the following table for the equation 2x – 3y = 7.

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

The two graph lines intersect at B(2, -1).
?x = 2 and y = -1 are the solutions of the given system of equations.

3. Solve the system of equations graphically:
2x + 3y = 8,
x – 2y + 3 = 0
Sol:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and
y-axis, respectively.
Graph of 2x + 3y = 8
2x + 3y = 8
?3y = (8 – 2x)
?y =
?? - ????
??           …(i)
Putting x = 1, we get y = 2.
Putting x = -5, we get y = 6.
Putting x = 7, we get y = -2.
Thus, we have the following table for the equation 2x + 3y = 8.
x 1 -5 7
y 2 6 -2
Now, plot the points A(1, 2), B(5, -6) and C(7, -2) on the graph paper.
Join AB and AC to get the graph line BC. Extend it on both ways.
Thus, BC is the graph of 2x + 3y = 8.

Graph of x - 2y + 3 = 0
x – 2y + 3 = 0
? 2y = (x + 3)
? y =
?? + ?? ??            …(ii)
Putting x = 1, we get y = 2.
Putting x = 3, we get y = 3.
Putting x = -3, we get y = 0.
Thus, we have the following table for the equation x – 2y + 3 = 0.
x 1 3 -3
y 2 3 0
Now, plot the points P (3, 3) and Q (-3, 0). The point A (1, 2) has already been plotted. Join
AP and QA and extend it on both ways.
Thus, PQ is the graph of x – 2y + 3 = 0.

The two graph lines intersect at A (1, 2).
? x = 1 and y = 2.

4. Solve the system of equations graphically:
2x - 5y + 4 = 0,
2x + y - 8 = 0
Sol:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and
y-axis, respectively.
Graph of 2x - 5y + 4 = 0
2x – 5y + 4 = 0
?5y = (2x + 4)
?y =
???? + ?? ??           …(i)

Putting x = -2, we get y = 0.
Putting x = 3, we get y = 2.
Putting x = 8, we get y = 4.
Thus, we have the following table for the equation 2x - 5y + 4 = 0.
x -2 3 8
y 0 2 4
Now, 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 on both ways.
Thus, AC is the graph of 2x - 5y + 4 = 0.

Graph of 2x + y - 8 = 0
2x + y - 8 = 0
? y = (8 – 2x)        …(ii)
Putting x = 1, we get y = 6.
Putting x = 3, we get y = 2.
Putting x = 2, we get y = 4.
Thus, we have the following table for the equation 2x + y - 8 = 0.
x 1 3 2
y 6 2 4
Now, plot the points P (1, 6) and Q (2, 4). The point B (3, 2) has already been plotted. Join
PQ and QB and extend it on both ways.
Thus, PB is the graph of 2x + y - 8 = 0.

The two graph lines intersect at B (3, 2).
?x = 3 and y = 2

5. Solve the system of equations graphically:
3x + 2y = 12,
5x – 2y = 4
Sol:
The given equations are:
3x + 2y = 12                       …..(i)
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## Mathematics (Maths) Class 10

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