CBSE Class 10 > Class 10 Notes > Mathematics (Maths) > RS Aggarwal Solutions: Linear Equations in two variables (Exercise 3A)
RS Aggarwal Solutions: Linear Equations in two variables (Exercise 3A)
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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)
Read MoreFAQs on RS Aggarwal Solutions: Linear Equations in two variables (Exercise 3A)
| 1. How do I solve linear equations in two variables using the substitution method? |  |
Ans. The substitution method involves expressing one variable in terms of the other from one equation, then substituting this expression into the second equation. This eliminates one variable, allowing you to solve for the remaining variable. Once found, substitute back to find the second variable's value. This algebraic technique is fundamental for solving simultaneous linear equations in two variables efficiently.
| 2. What's the difference between consistent and inconsistent linear equations in two variables? | |
Ans. Consistent equations have at least one solution where both equations are satisfied simultaneously, while inconsistent equations have no common solution. When graphed, consistent lines intersect or overlap; inconsistent lines remain parallel. Understanding this distinction helps identify whether a system of linear equations in two variables will yield valid answers or prove impossible to solve.
| 3. Why do I get different answers when solving the same pair of equations using elimination versus substitution? | |
Ans. Both elimination and substitution methods yield identical answers when applied correctly to the same linear equations in two variables. If answers differ, a computational error occurred during the solving process. Cross-checking your arithmetic in each step-whether multiplying coefficients or combining terms-reveals mistakes. Both methods are equally valid for solving simultaneous equations.
| 4. How do I check if my solution to a system of linear equations in two variables is correct? | |
Ans. Substitute both variable values back into both original equations. If both equations are satisfied (left side equals right side), your solution is correct. This verification step prevents errors from going undetected and builds confidence in your answer. It's essential practice for board exam preparation and ensures accuracy in solving linear equations problems.
| 5. When should I use the graphical method instead of algebraic methods for linear equations in two variables? | |
Ans. The graphical method works best for visualising solutions and understanding geometric interpretations-where lines intersect represents the solution point. However, for precise numerical answers in CBSE exams, algebraic methods (substitution or elimination) are preferred as they avoid measurement errors. Use graphical representation to verify algebraic solutions or when understanding the relationship between equations matters more than exact values.
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