Class 9 Exam > Class 9 Notes > Mathematics (Maths) Class 9 > RS Aggarwal Solutions: Linear Equations in Two Variables
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Page 1
Q u e s t i o n : 1
Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case.
i
3x + 5y = 7.5
ii
2x -
y
5
+6 = 0
iii
3y – 2x = 6
iv
4x = 5y
v
x
5
-
y
6
= 1
vi
v
2x +
v
3y = 5
S o l u t i o n :
i
3x + 5y = 7.5
This can be expressed in the form ax + by + c = 0 as 3x +5y +(-7. 5) = 0
.
ii
2x -
y
5
+6 = 0
This can be expressed in the form ax + by + c = 0 as 2x + -
1
5
y +6 = 0
iii
3y – 2x = 6
This can be expressed in the form ax + by + c = 0 as 2x +(-3y)+6 = 0
.
iv
4x = 5y
This can be expressed in the form ax + by + c = 0 as 4x -5y +0 = 0
v
x
5
-
y
6
= 1
This can be expressed in the form ax + by + c = 0 as 6x -5y = 30
vi
v
2x +
v
3y = 5
This can be expressed in the form ax + by + c = 0 as
v
2x +
v
3y +(-5) = 0
Q u e s t i o n : 2
Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case.
i
x = 6
ii
3x – y = x – 1
iii
2x + 9 = 0
iv
4y = 7
v
x + y = 4
vi
x
2
-
y
3
=
1
6
+y
( )
Page 2
Q u e s t i o n : 1
Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case.
i
3x + 5y = 7.5
ii
2x -
y
5
+6 = 0
iii
3y – 2x = 6
iv
4x = 5y
v
x
5
-
y
6
= 1
vi
v
2x +
v
3y = 5
S o l u t i o n :
i
3x + 5y = 7.5
This can be expressed in the form ax + by + c = 0 as 3x +5y +(-7. 5) = 0
.
ii
2x -
y
5
+6 = 0
This can be expressed in the form ax + by + c = 0 as 2x + -
1
5
y +6 = 0
iii
3y – 2x = 6
This can be expressed in the form ax + by + c = 0 as 2x +(-3y)+6 = 0
.
iv
4x = 5y
This can be expressed in the form ax + by + c = 0 as 4x -5y +0 = 0
v
x
5
-
y
6
= 1
This can be expressed in the form ax + by + c = 0 as 6x -5y = 30
vi
v
2x +
v
3y = 5
This can be expressed in the form ax + by + c = 0 as
v
2x +
v
3y +(-5) = 0
Q u e s t i o n : 2
Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case.
i
x = 6
ii
3x – y = x – 1
iii
2x + 9 = 0
iv
4y = 7
v
x + y = 4
vi
x
2
-
y
3
=
1
6
+y
( )
S o l u t i o n :
i
x = 6
In the form of ax + by + c = 0 we have x +0y +(-6) = 0
where a = 1, b = 0 and c = -6
.
ii
3x – y = x – 1
In the form of ax + by + c = 0 we have 2x +(-1y)+1 = 0
where a = 2, b = -1
and c = 1
iii
2x + 9 = 0
In the form of ax + by + c = 0 we have 2x +(-1y)+1 = 0
where a = 2, b = -1
and c = 1
iv
4y = 7
In the form of ax + by + c = 0 we have 0x +4y +(-7) = 0
where a = 0, b = 4 and c = -7
v
x + y = 4
In the form of ax + by + c = 0 we have 1x +1y +(-4) = 0
where a = 1, b = 1, c = -4
vi
x
2
-
y
3
=
1
6
+y
In the form of ax + by + c = 0 we have 3x -2y = 1 +6y ? 3x -8y +(-1) = 0
where a = 3, b = -8, c = -1
.
Q u e s t i o n : 3
Check which of the following are the solutions of the equation 5x – 4y = 20.
i
4, 0
ii
0, 5
iii
-2,
5
2
iv
0, – 5
v
2,
-5
2
S o l u t i o n :
The equation given is 5x – 4y = 20.
i
4, 0
Putting the value in the given equation we have
LHS: 5(4)-4(0) = 20RHS: 20LHS = RHS
Thus, 4, 0
is a solution of the given equation.
ii
0, 5
Putting the value in the given equation we have
LHS: 5(0)-4(5) = 0 -20 = -20RHS: 20LHS ? RHS
( )
( )
Page 3
Q u e s t i o n : 1
Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case.
i
3x + 5y = 7.5
ii
2x -
y
5
+6 = 0
iii
3y – 2x = 6
iv
4x = 5y
v
x
5
-
y
6
= 1
vi
v
2x +
v
3y = 5
S o l u t i o n :
i
3x + 5y = 7.5
This can be expressed in the form ax + by + c = 0 as 3x +5y +(-7. 5) = 0
.
ii
2x -
y
5
+6 = 0
This can be expressed in the form ax + by + c = 0 as 2x + -
1
5
y +6 = 0
iii
3y – 2x = 6
This can be expressed in the form ax + by + c = 0 as 2x +(-3y)+6 = 0
.
iv
4x = 5y
This can be expressed in the form ax + by + c = 0 as 4x -5y +0 = 0
v
x
5
-
y
6
= 1
This can be expressed in the form ax + by + c = 0 as 6x -5y = 30
vi
v
2x +
v
3y = 5
This can be expressed in the form ax + by + c = 0 as
v
2x +
v
3y +(-5) = 0
Q u e s t i o n : 2
Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case.
i
x = 6
ii
3x – y = x – 1
iii
2x + 9 = 0
iv
4y = 7
v
x + y = 4
vi
x
2
-
y
3
=
1
6
+y
( )
S o l u t i o n :
i
x = 6
In the form of ax + by + c = 0 we have x +0y +(-6) = 0
where a = 1, b = 0 and c = -6
.
ii
3x – y = x – 1
In the form of ax + by + c = 0 we have 2x +(-1y)+1 = 0
where a = 2, b = -1
and c = 1
iii
2x + 9 = 0
In the form of ax + by + c = 0 we have 2x +(-1y)+1 = 0
where a = 2, b = -1
and c = 1
iv
4y = 7
In the form of ax + by + c = 0 we have 0x +4y +(-7) = 0
where a = 0, b = 4 and c = -7
v
x + y = 4
In the form of ax + by + c = 0 we have 1x +1y +(-4) = 0
where a = 1, b = 1, c = -4
vi
x
2
-
y
3
=
1
6
+y
In the form of ax + by + c = 0 we have 3x -2y = 1 +6y ? 3x -8y +(-1) = 0
where a = 3, b = -8, c = -1
.
Q u e s t i o n : 3
Check which of the following are the solutions of the equation 5x – 4y = 20.
i
4, 0
ii
0, 5
iii
-2,
5
2
iv
0, – 5
v
2,
-5
2
S o l u t i o n :
The equation given is 5x – 4y = 20.
i
4, 0
Putting the value in the given equation we have
LHS: 5(4)-4(0) = 20RHS: 20LHS = RHS
Thus, 4, 0
is a solution of the given equation.
ii
0, 5
Putting the value in the given equation we have
LHS: 5(0)-4(5) = 0 -20 = -20RHS: 20LHS ? RHS
( )
( )
Thus, 0, 5
is not a solution of the given equation.
iii
-2,
5
2
Putting the value in the given equation we have
LHS: 5(-2)-4
5
2
= -10 -10 = -20RHS: 20LHS ? RHS
Thus, -2,
5
2
is not a solution of the given equation.
iv
0, – 5
Putting the value in the given equation we have
LHS: 5(0)-4(-5) = 0 +20 = 20RHS: 20LHS = RHS
Thus, 0, – 5
is a solution of the given equation.
v
2,
-5
2
Putting the value in the given equation we have
LHS: 5(2)-4
-5
2
= 10 +10 = 20RHS: 20LHS = RHS
Thus, 2,
-5
2
is a solution of the given equation.
Q u e s t i o n : 4
Find five different solutions of each of the following equations:
i
2x – 3y = 6
ii
2x
5
+
3y
10
= 3
iii
3y = 4x
S o l u t i o n :
i
2x – 3y = 6
x 0 3
-
3
9
2
2
y
-
2
0
-
4
1
-2
3
ii
( )
( )
( )
( )
( )
( )
Page 4
Q u e s t i o n : 1
Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case.
i
3x + 5y = 7.5
ii
2x -
y
5
+6 = 0
iii
3y – 2x = 6
iv
4x = 5y
v
x
5
-
y
6
= 1
vi
v
2x +
v
3y = 5
S o l u t i o n :
i
3x + 5y = 7.5
This can be expressed in the form ax + by + c = 0 as 3x +5y +(-7. 5) = 0
.
ii
2x -
y
5
+6 = 0
This can be expressed in the form ax + by + c = 0 as 2x + -
1
5
y +6 = 0
iii
3y – 2x = 6
This can be expressed in the form ax + by + c = 0 as 2x +(-3y)+6 = 0
.
iv
4x = 5y
This can be expressed in the form ax + by + c = 0 as 4x -5y +0 = 0
v
x
5
-
y
6
= 1
This can be expressed in the form ax + by + c = 0 as 6x -5y = 30
vi
v
2x +
v
3y = 5
This can be expressed in the form ax + by + c = 0 as
v
2x +
v
3y +(-5) = 0
Q u e s t i o n : 2
Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case.
i
x = 6
ii
3x – y = x – 1
iii
2x + 9 = 0
iv
4y = 7
v
x + y = 4
vi
x
2
-
y
3
=
1
6
+y
( )
S o l u t i o n :
i
x = 6
In the form of ax + by + c = 0 we have x +0y +(-6) = 0
where a = 1, b = 0 and c = -6
.
ii
3x – y = x – 1
In the form of ax + by + c = 0 we have 2x +(-1y)+1 = 0
where a = 2, b = -1
and c = 1
iii
2x + 9 = 0
In the form of ax + by + c = 0 we have 2x +(-1y)+1 = 0
where a = 2, b = -1
and c = 1
iv
4y = 7
In the form of ax + by + c = 0 we have 0x +4y +(-7) = 0
where a = 0, b = 4 and c = -7
v
x + y = 4
In the form of ax + by + c = 0 we have 1x +1y +(-4) = 0
where a = 1, b = 1, c = -4
vi
x
2
-
y
3
=
1
6
+y
In the form of ax + by + c = 0 we have 3x -2y = 1 +6y ? 3x -8y +(-1) = 0
where a = 3, b = -8, c = -1
.
Q u e s t i o n : 3
Check which of the following are the solutions of the equation 5x – 4y = 20.
i
4, 0
ii
0, 5
iii
-2,
5
2
iv
0, – 5
v
2,
-5
2
S o l u t i o n :
The equation given is 5x – 4y = 20.
i
4, 0
Putting the value in the given equation we have
LHS: 5(4)-4(0) = 20RHS: 20LHS = RHS
Thus, 4, 0
is a solution of the given equation.
ii
0, 5
Putting the value in the given equation we have
LHS: 5(0)-4(5) = 0 -20 = -20RHS: 20LHS ? RHS
( )
( )
Thus, 0, 5
is not a solution of the given equation.
iii
-2,
5
2
Putting the value in the given equation we have
LHS: 5(-2)-4
5
2
= -10 -10 = -20RHS: 20LHS ? RHS
Thus, -2,
5
2
is not a solution of the given equation.
iv
0, – 5
Putting the value in the given equation we have
LHS: 5(0)-4(-5) = 0 +20 = 20RHS: 20LHS = RHS
Thus, 0, – 5
is a solution of the given equation.
v
2,
-5
2
Putting the value in the given equation we have
LHS: 5(2)-4
-5
2
= 10 +10 = 20RHS: 20LHS = RHS
Thus, 2,
-5
2
is a solution of the given equation.
Q u e s t i o n : 4
Find five different solutions of each of the following equations:
i
2x – 3y = 6
ii
2x
5
+
3y
10
= 3
iii
3y = 4x
S o l u t i o n :
i
2x – 3y = 6
x 0 3
-
3
9
2
2
y
-
2
0
-
4
1
-2
3
ii
( )
( )
( )
( )
( )
( )
2x
5
+
3y
10
= 3
x 0
15
2
5 10 3
y 10 0
10
3
-10
3
6
iii
3y = 4x
x 3
-
3
-
6
6 0
y 4
-
4
-
8
8 0
Q u e s t i o n : 5
If x = 3 and y = 4 is a solution of the equation 5x – 3y = k, find the value of k.
S o l u t i o n :
Given: 5x – 3y = k
Since x = 3 and y = 4 is a solution of the given equation so, it should satisfy the equation.
5(3)-3(4) = k ? 15 -12 = k ? 3 = k
Q u e s t i o n : 6
If x = 3k + 2 and y = 2k – 1 is a solution of the equation 4x – 3y + 1 = 0, find the value of k.
S o l u t i o n :
Given: 4x – 3y + 1 = 0 .....1
x = 3k + 2 and y = 2k – 1
Putting these values in the equation 1
we get
4(3k +2)-3(2k -1)+1 = 0 ? 12k +8 -6k +3 +1 = 0 ? 6k +12 = 0 ? k +2 = 0 ? k = -2
Q u e s t i o n : 7
The cost of 5 pencils is equal to the cost of 2 ballpoints. Write a linear equation in two variables to represent this statement. (Take
the cost of a pencil to be x and that of a ballpoint to be y).
S o l u t i o n :
Let cost of a pencil to be x and that of a ballpoint to be y.
Cost of 5 pencils = 5x
Cost of 2 ballpoints = 2y
Cost of 5 pencils = cost of 2 ballpoints
? 5x = 2y ? 5x -2y = 0
Page 5
Q u e s t i o n : 1
Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case.
i
3x + 5y = 7.5
ii
2x -
y
5
+6 = 0
iii
3y – 2x = 6
iv
4x = 5y
v
x
5
-
y
6
= 1
vi
v
2x +
v
3y = 5
S o l u t i o n :
i
3x + 5y = 7.5
This can be expressed in the form ax + by + c = 0 as 3x +5y +(-7. 5) = 0
.
ii
2x -
y
5
+6 = 0
This can be expressed in the form ax + by + c = 0 as 2x + -
1
5
y +6 = 0
iii
3y – 2x = 6
This can be expressed in the form ax + by + c = 0 as 2x +(-3y)+6 = 0
.
iv
4x = 5y
This can be expressed in the form ax + by + c = 0 as 4x -5y +0 = 0
v
x
5
-
y
6
= 1
This can be expressed in the form ax + by + c = 0 as 6x -5y = 30
vi
v
2x +
v
3y = 5
This can be expressed in the form ax + by + c = 0 as
v
2x +
v
3y +(-5) = 0
Q u e s t i o n : 2
Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case.
i
x = 6
ii
3x – y = x – 1
iii
2x + 9 = 0
iv
4y = 7
v
x + y = 4
vi
x
2
-
y
3
=
1
6
+y
( )
S o l u t i o n :
i
x = 6
In the form of ax + by + c = 0 we have x +0y +(-6) = 0
where a = 1, b = 0 and c = -6
.
ii
3x – y = x – 1
In the form of ax + by + c = 0 we have 2x +(-1y)+1 = 0
where a = 2, b = -1
and c = 1
iii
2x + 9 = 0
In the form of ax + by + c = 0 we have 2x +(-1y)+1 = 0
where a = 2, b = -1
and c = 1
iv
4y = 7
In the form of ax + by + c = 0 we have 0x +4y +(-7) = 0
where a = 0, b = 4 and c = -7
v
x + y = 4
In the form of ax + by + c = 0 we have 1x +1y +(-4) = 0
where a = 1, b = 1, c = -4
vi
x
2
-
y
3
=
1
6
+y
In the form of ax + by + c = 0 we have 3x -2y = 1 +6y ? 3x -8y +(-1) = 0
where a = 3, b = -8, c = -1
.
Q u e s t i o n : 3
Check which of the following are the solutions of the equation 5x – 4y = 20.
i
4, 0
ii
0, 5
iii
-2,
5
2
iv
0, – 5
v
2,
-5
2
S o l u t i o n :
The equation given is 5x – 4y = 20.
i
4, 0
Putting the value in the given equation we have
LHS: 5(4)-4(0) = 20RHS: 20LHS = RHS
Thus, 4, 0
is a solution of the given equation.
ii
0, 5
Putting the value in the given equation we have
LHS: 5(0)-4(5) = 0 -20 = -20RHS: 20LHS ? RHS
( )
( )
Thus, 0, 5
is not a solution of the given equation.
iii
-2,
5
2
Putting the value in the given equation we have
LHS: 5(-2)-4
5
2
= -10 -10 = -20RHS: 20LHS ? RHS
Thus, -2,
5
2
is not a solution of the given equation.
iv
0, – 5
Putting the value in the given equation we have
LHS: 5(0)-4(-5) = 0 +20 = 20RHS: 20LHS = RHS
Thus, 0, – 5
is a solution of the given equation.
v
2,
-5
2
Putting the value in the given equation we have
LHS: 5(2)-4
-5
2
= 10 +10 = 20RHS: 20LHS = RHS
Thus, 2,
-5
2
is a solution of the given equation.
Q u e s t i o n : 4
Find five different solutions of each of the following equations:
i
2x – 3y = 6
ii
2x
5
+
3y
10
= 3
iii
3y = 4x
S o l u t i o n :
i
2x – 3y = 6
x 0 3
-
3
9
2
2
y
-
2
0
-
4
1
-2
3
ii
( )
( )
( )
( )
( )
( )
2x
5
+
3y
10
= 3
x 0
15
2
5 10 3
y 10 0
10
3
-10
3
6
iii
3y = 4x
x 3
-
3
-
6
6 0
y 4
-
4
-
8
8 0
Q u e s t i o n : 5
If x = 3 and y = 4 is a solution of the equation 5x – 3y = k, find the value of k.
S o l u t i o n :
Given: 5x – 3y = k
Since x = 3 and y = 4 is a solution of the given equation so, it should satisfy the equation.
5(3)-3(4) = k ? 15 -12 = k ? 3 = k
Q u e s t i o n : 6
If x = 3k + 2 and y = 2k – 1 is a solution of the equation 4x – 3y + 1 = 0, find the value of k.
S o l u t i o n :
Given: 4x – 3y + 1 = 0 .....1
x = 3k + 2 and y = 2k – 1
Putting these values in the equation 1
we get
4(3k +2)-3(2k -1)+1 = 0 ? 12k +8 -6k +3 +1 = 0 ? 6k +12 = 0 ? k +2 = 0 ? k = -2
Q u e s t i o n : 7
The cost of 5 pencils is equal to the cost of 2 ballpoints. Write a linear equation in two variables to represent this statement. (Take
the cost of a pencil to be x and that of a ballpoint to be y).
S o l u t i o n :
Let cost of a pencil to be x and that of a ballpoint to be y.
Cost of 5 pencils = 5x
Cost of 2 ballpoints = 2y
Cost of 5 pencils = cost of 2 ballpoints
? 5x = 2y ? 5x -2y = 0
Q u e s t i o n : 8
Draw the graph of each of the following equations.
i
x = 4
ii
x + 4 = 0
iii
y = 3
iv
y = –3
v
x = –2
vi
x = 5
vii
y + 5 = 0
viii
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FAQs on RS Aggarwal Solutions: Linear Equations in Two Variables - Mathematics (Maths) Class 9
| 1. What are linear equations in two variables? |  |
Ans. Linear equations in two variables are algebraic expressions that involve two variables and have a degree of 1. These equations can be represented by straight lines on a coordinate plane and have the general form of ax + by = c, where a, b, and c are constants.
| 2. How do you solve a linear equation in two variables? | |
Ans. To solve a linear equation in two variables, we need to find the values of the variables that make the equation true. This can be done by using various methods such as substitution, elimination, or graphing. By substituting the value of one variable from one equation into the other equation, we can find the value of the other variable.
| 3. How is graphing used to solve linear equations in two variables? | |
Ans. Graphing is a graphical method used to solve linear equations in two variables. By plotting the given equations on a coordinate plane, the point of intersection of the lines represents the solution to the equations. This method is useful when the equations are given in a graphical format or when we want to visually understand the solution.
| 4. What is the importance of linear equations in two variables? | |
Ans. Linear equations in two variables are important in various fields such as physics, economics, engineering, and computer science. They help in modeling real-life situations, analyzing relationships between variables, and making predictions. These equations provide a fundamental understanding of the concept of lines and their intersections, which is crucial in many mathematical applications.
| 5. Can linear equations in two variables have infinitely many solutions? | |
Ans. Yes, linear equations in two variables can have infinitely many solutions. This occurs when the two equations are equivalent or represent the same line. In such cases, every point on the line is a solution to the equations. This concept is known as dependent equations, where the equations are not distinct and have an infinite number of solutions.
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Jan 01, 2025 Last updated |
Document Description: RS Aggarwal Solutions: Linear Equations in Two Variables for Class 9 2025 is part of Mathematics (Maths) Class 9 preparation.
The notes and questions for RS Aggarwal Solutions: Linear Equations in Two Variables have been prepared according to the Class 9 exam syllabus. Information about RS Aggarwal Solutions: Linear Equations in Two Variables covers topics
like and RS Aggarwal Solutions: Linear Equations in Two Variables Example, for Class 9 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RS Aggarwal Solutions: Linear Equations in Two Variables.
Introduction of RS Aggarwal Solutions: Linear Equations in Two Variables in English is available as part of our Mathematics (Maths) Class 9
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Full syllabus notes, lecture & questions for RS Aggarwal Solutions: Linear Equations in Two Variables | Mathematics (Maths) Class 9 - Class 9 | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 9 | Best notes, free PDF download
Information about RS Aggarwal Solutions: Linear Equations in Two Variables
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Additional Information about RS Aggarwal Solutions: Linear Equations in Two Variables for Class 9 Preparation
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The content of RS Aggarwal Solutions: Linear Equations in Two Variables is prepared as per the latest Class 9 syllabus.
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