RS Aggarwal Solutions Class 9 Maths Chapter 4 Linear Equations in Two Variables (Updated For 2024)

RS Aggarwal Solutions Class 9 Maths Chapter 4 Linear Equations in Two Variables: Class 9 Maths is made easier with the RS Aggarwal Solutions Class 9 Maths. Be it exam preparation or class assignments, RS Aggarwal Solutions Class 9 Maths Chapter 4 Linear Equations in Two Variables has got you covered. The solutions are well-written, well-explained, and reliable.

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RS Aggarwal Solutions Class 9 Maths Chapter 4 Linear Equations in Two Variables

Access RS Aggarwal Solutions Class 9 Maths Chapter 4 Linear Equations in Two Variables

Question 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
⇒3x+5y=15/2
⇒6x+10y-15= 0 [Multiplying by 2 both side]
On comparing we have, a=6,b=10 and c=-15

(ii) 2x- (y/5) + 6=0
⇒10x – y+30 = 0 [Multiplying by 5 both side]
On comparing we have, a=10, b= – 1 and c=30

(iii) 3y-2x=6
⇒-2x+3y-6=0
On comparing we have, a=-2, b=3 and c=- 6
(iv) 4x=5y
⇒4x-5y+0=0
On comparing we have, a= 4,b=- 5 and c= 0

⇒ 6x-5y=30
⇒ 6x-5y-30 =0
On comparing we have, a=6,b=-5 and c=-30

(vi) √2 x+√3 y=5
⇒√2 x+√3 y-5=0
On comparing we have, a=√2,b=√3 and c=-5

Question 2:

Express each of the following equations in the form of ax+by+c=0 and indicate the values of a, b, c in each case:

(i) x = 6
⇒ x + 0.y – 6 = 0
On comparing we have, a=1,b=0 and c=- 6

(ii) 3x-y=x-1
⇒ 3x-y-x+1=0
⇒2x-y+1=0
On comparing we have, a=2,b=-1 and c=1

(iii) 2x+9=0
⇒2x+0.y+9=0
On comparing we have, a=2,b=0 and c=9

(iv) 4y=7
⇒ 0.x+4y-7=0
On comparing we have, a=0,b=4 and c=-7

(v) x+y = 4
⇒ x+y- 4=0
On comparing we have, a=1, b=1 and c=-4

⇒3x-2y=1+6y
⇒3x-2y-1-6y=0
⇒3x- 8y-1=0
On comparing we have, a=3, b=- 8 and c=-1

Question 3:

Check which of the following are the solutions of the equation 5x-4y=20.
(i) (4, 0)
Put x = 4 and y = 0 in 5x-4y=20.
5(4)-4(0)=20.
20=20 L.H.S = R.H.S
Therefore (4, 0) is a solution of the equation.

(ii) (0, 5)
Put x = 0 and y = 5 in 5x-4y=20.
5(0)-4(5)=20.
-20≠20 L.H.S ≠ R.H.S
Therefore (0, 5) is not a solution of the equation.

(iii) (-2, 5/2)
Put x = – 2 and y = 5/2 in 5x-4y=20.
5(-2)-4(5/2)=20.
-20≠20 L.H.S ≠ R.H.S
Therefore (-2, 5/2) is not a solution of the equation.

(iv) (0,-5)
Put x = 0 and y = – 5 in 5x-4y=20.
5(0)-4(-5)=20.
20=20 L.H.S = R.H.S
Therefore (0, – 5) is a solution of the equation.

(v) (2,-5/2)
Put x = 2 and y = -5/2 in 5x-4y=20.
5(2)-4(-5/2)=20.
20=20 L.H.S = R.H.S
Therefore (2,-5/2) is a solution of the equation.

From above (4, 0), (0, – 5), and (2,-5/2) are the solutions.

Question 4:

Find five different solutions of each of the following equations:

Solution:

Question 5:

If x = 3 and y = 4 is a solution of the equation 5x-3y=k, find value of k.

Solution:

Putting x = 3 and y = 4 in 5x-3y=k , we get
⇒ 5(3)-3(4)=k
⇒ k=15-12=3
Hence, k = 3

Question 6:

If x=3k+2 and y=2k-1 is a solution of the equation 4x-3y+1=0, find the value of k.

Solution:

Putting the value of x=3k+2 and y=2k-1 in 4x-3y+1=0, we get
⇒ 4(3k+2)-3(2k-1)+1=0
⇒ 12k+8-6k+3+1=0
⇒ 6k+12=0 i.e. k=-2

Question 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 the pencil to be Rs. x and that of a ballpoint to be Rs. y).

Solution:

Let cost of pencil is x and cost of ballpoint is y
Then cost of 5 pencil = 5x
And cost of 2 ballpoint = 2y
According to question, 5x = 2y
i.e. 5x – 2y = 0

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