Class 10 Exam  >  Class 10 Notes  >  Mathematics (Maths) Class 10  >  RS Aggarwal Solutions: Linear Equations in two variables (Exercise 3A)

RS Aggarwal Solutions: Linear Equations in two variables (Exercise 3A) | Mathematics (Maths) Class 10 PDF Download

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 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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FAQs on RS Aggarwal Solutions: Linear Equations in two variables (Exercise 3A) - Mathematics (Maths) Class 10

1. How do you solve a linear equation in two variables?
Ans. To solve a linear equation in two variables, you need to isolate one variable in terms of the other variable and substitute it back into the equation to find the value of the remaining variable. This can be done using various methods such as substitution method, elimination method, or graphical method.
2. What is the importance of solving linear equations in two variables?
Ans. Solving linear equations in two variables is important as it helps in finding the relationship between two variables and determining their values. This is particularly useful in real-life situations where multiple variables are involved, such as in economics, physics, or engineering problems.
3. Can a linear equation in two variables have more than one solution?
Ans. Yes, a linear equation in two variables can have more than one solution. This happens when the equation represents a line, and the line intersects with another line at more than one point. In such cases, the variables can take on different values that satisfy the equation.
4. What is the difference between consistent and inconsistent linear equations in two variables?
Ans. In a consistent linear equation in two variables, the equation has at least one solution, and the lines represented by the equation intersect at a point. On the other hand, in an inconsistent linear equation, the lines represented by the equation are parallel and do not intersect, resulting in no solution.
5. How can linear equations in two variables be represented graphically?
Ans. Linear equations in two variables can be represented graphically by plotting the points that satisfy the equation on a coordinate plane. Each point represents a solution to the equation, and when these points are connected, they form a straight line. The slope-intercept form (y = mx + b) or the general form (Ax + By = C) of the equation can be used to determine the slope and y-intercept of the line, which further helps in graphing the equation.
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