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

Mathematics (Maths) Class 10

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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