Class 9 Exam  >  Class 9 Notes  >  RD Sharma Solutions for Class 9 Mathematics  >  RD Sharma Solutions Ex-13.2, Linear Equation In Two Variables, Class 9, Maths

RD Sharma Solutions Ex-13.2, Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics PDF Download

Q 1 : Write two solutions for each of the following equations:

(i) 5x – 2y = 7

(ii) x = 6y

(iii) x +πy = 4

(iv) 2/3x – y = 4.

A 1 :

(i) We are given,

3x + 4y = 7

Substituting x = 1

In the given equation,

We get

3 x1 + 4y  = 7

4y = 7 – 3  4 = 4Y

Y = I

Thus x = 1 and y = 1 is the solution of 3x + 4y = 7

Substituting x = 2 in the given equation,

we get 3×2 + 4y = 7

4y = 7 – 6

y = 1/4

Thus x = 2 and  y = 1/4 is the solution of 3x + 4y = 7

(ii) We are given, x = 6y

Substituting x =0 in the given equation,

we get 0 = 6y

6 y = 0

y = 0

Thus x = 0,       ⇒ Solution (0,0)

Substituting x=6

6 = 6y

y =  6/6

y = 1          ⇒ Solution (6,1)

(iii)  We are given x+πy = 4

Substituting x = 0 in the given equation,

We get 0+πy = 4

πy = 4

RD Sharma Solutions Ex-13.2, Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

⇒ Solution  = RD Sharma Solutions Ex-13.2, Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Substituting y = 0 in the given equation, we get

x + 0 = 4

x = 4

⇒ Solution = (4, 0)

(iv) We are given RD Sharma Solutions Ex-13.2, Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Substituting x = 0 in the given equation, we get

0 – y = 4

y = – 4

Thus x = 0 and y = – 4  is a solution

Substituting x = 3 in the given equation, we get

RD Sharma Solutions Ex-13.2, Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

2 – y = 4

y = 2 – 4

y = -2

Thus x = 3 and y = -2 is a solution


Q 2 : Write two solutions of the form x = 0,  y = a and x = b, y = 0 for each of the following  equations :

(i) 5x – 2y =10
(ii) -4x + 3y =12
(iii) 2x + 3y = 24

A 2 :      

(i) We are given,

5x – 2y = 10

Substituting x = 0 in the given equation,

We get;

5×0 – 2y = 10

– 2y = 10

– y = 10/2

y = – 5

Thus x = 0 and y = -5 is the solution of 5x-2y = 10

Substituting y = 0 in the given equation, we get 5x —2 x 0 = 10

5x = 10

x = 10/2

x = 2

Thus x = 2 and y = 0 is a solution of 5x-2y = 10

(ii) We are given, – 4x + 3y = 12

Substituting x = 0 in the given equation,

we get;

-4×0 + 3y = 12

3y = 12

y = 4

Thus x = 0 and y = 4 is a solution of the -4x + 3y = 12

Substituting y = 0 in the given equation, we get;

-4 x + 3 x 0 = 12

– 4x = 12

x =-12/4

x = -3

Thus x = -3 and y =0 is a solution of -4x + 3y = 12

(iii) We are given, 2x + 3y = 24

Substituting x = 0 in the given equation, we get;

2 x 0 + 3y = 24

3y =24

y = 24/3

y = 8

Thus x = 0 and y = 8 is a solution of 2x+ 3y = 24

Substituting y = 0 in the given equation, we get;

2x +3 x 0 = 24

2x = 24

x = 24/2

x =12

Thus x =12 and y = 0 is a solution of 2x + 3y = 24


Q3: Check which of the following are solutions of the equation 2x — y = 6 and Which are not :

(i) (3 , 0)
(ii) (0 , 6)
(iii) (2 , – 2) 
(iv) RD Sharma Solutions Ex-13.2, Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics
(v)RD Sharma Solutions Ex-13.2, Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

A3:

We are given, 2x – y = 6

(i) In the equation 2 x – y = 6,

We have L.H.S = 2x — y and R.H.S = 6

Substituting x = 3 and y = 0 in 2x – y ,

We get L.H.S = 2 x 3 – 0 = 6

⇒ L.H.S = R.H.S

⇒ (3,0) is a solution of 2x — y = 6.

(ii) In the equation 2x — y = 6,

We have L.H.S = 2x— y and R.H.S = 6

Substituting x = 0 and y = 6 in 2x — y

We get L.H.S = 2 x 0 – 6 = – 6

⇒ L.H.S ≠ R.H.S

⇒ (0,6) is not a solution of 2x — y = 6.

(iii) In the equation 2x — y = 6,

We have L.H.S = 2x – y and R.H.S = 6

Substituting x = 2 and y = – 2 in 2x — y,

We get L.H.S = 2 x 2 – (-2) = 6

⇒ L.H.S = R.H.S

⇒ (2,-2) is a solution of 2x — y = 6.

(iv) In the equation 2x — y = 6,

We have L.H.S = 2x- y and R.H.S=6

Substituting x = √3 and y = 0 in 2x — y,

We get L.H.S = 2 x √3 – 0

⇒ L.H.S ≠ R.H.S

⇒(√3,0)is not a solution of 2x — y = 6.

(v) In the equation 2x — y = 6,

We have L.H.S = 2x – y and R.H.S = 6

Substituting x = 1/2 and y =  in 2x — y, we get

L.H.S = 2 x(1/2) – (-5)

⇒1 + 5 =  6

⇒  L.H.S = R.H.S

RD Sharma Solutions Ex-13.2, Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics is a solution of 2x — y = 6.


Q4:  If x = -1, y = 2 is a solution of the equation 3x + 4y = k, find the value of k.

A4 :

We are given, 3 x + 4 y = k

Given that, (-1,2) is the solution of equation 3x + 4y = k.

Substituting x = -1 and y = 2 in 3x + 4y = k,

We get; 3x – 1 + 4 x 2 = k

K = – 3 + 8

k = 5


Q 5 : Find the value of λ, if x = –λ and y = 5/2 is a solution of the equation x + 4y -7 = 0

A 5 :

We are given,

x + 4y – 7 = 0

(-λ,- 5)is a solution of equation 3x + 4y = k

Substituting x = – λ and y = 5/2 in x + 4y – 7 = 0

RD Sharma Solutions Ex-13.2, Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

RD Sharma Solutions Ex-13.2, Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics
λ = 10 – 7

λ = 3


Q 6 : If x = 2a + 1 and y = a -1 is a solution of the equation 2x – 3y + 5 = 0, find the value of a.

A 6:

We are given, 2x —3y +5 = 0

(2a + 1, a – 1) is the solution of equation 2x – 3y + 5 = 0.

Substituting x = 2a + 1 and y = a – 1 in 2x – 3y + 5 = 0,

We get 2 x 2a + (1- 3) x a – 1 + 5 = 0

⇒ 4a + 2 – 3a + 3 + 5 = 0

⇒ a + 10 = 0

⇒ a = – 10


Q 7 : If x = 1 and y = 6 is a solution of the equation 8x – ay + a2 = 0, find the values of a.

A 7 :

We are given,

8x – ay + a2 = 0

(1 , 6) is a solution of equation 8x — ay + a2 =0

Substituting x = 1 and y = 6 in 8x – ay + a2 =  0 ,we get

8 x 1 – a x 6 + a2 = 0

⇒ a2 – 6a + 8 = 0

Using quadratic factorization

a2 – 4a – 2a + 8 = 0

a(a – 4) – 2 (a – 4) = 0

(a – 2) (a – 4) = 0

a = 2, 4

The document RD Sharma Solutions Ex-13.2, Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
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FAQs on RD Sharma Solutions Ex-13.2, Linear Equation In Two Variables, Class 9, Maths - RD Sharma Solutions for Class 9 Mathematics

1. What are linear equations in two variables?
Ans. Linear equations in two variables are algebraic equations that involve two variables, usually represented by x and y, and have a degree of 1. These equations can be written in the form ax + by = c, where a, b, and c are constants.
2. How do you solve linear equations in two variables?
Ans. To solve a linear equation in two variables, we need to find the values of both variables that satisfy the equation. This can be done by using various methods like substitution, elimination, or graphical representation. By substituting the value of one variable from one equation into the other equation, we can find the value of the remaining variable. Repeating this process will give us the values of both variables.
3. What is the importance of linear equations in two variables?
Ans. Linear equations in two variables have various applications in real-life situations. They are used to solve problems related to cost and revenue analysis, distance and speed calculations, optimization of resources, and many more. By finding the solutions to these equations, we can make informed decisions and solve practical problems efficiently.
4. Can linear equations in two variables have more than one solution?
Ans. Yes, linear equations in two variables can have more than one solution. It depends on the nature of the equations. If the equations are consistent and dependent, they will have infinitely many solutions. This means that any point on the line representing the equation will be a solution. On the other hand, if the equations are consistent and independent, they will have a unique solution, which is the point of intersection of the lines representing the equations.
5. How are linear equations in two variables 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 equation can be represented by a straight line. The point of intersection of these lines represents the solution to the system of equations. If the lines are parallel, it indicates that the equations have no common solution. If the lines coincide, it means that the equations have infinitely many solutions.
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