SEBA Class 9 Maths Chapter 4 Linear Equations in Two Variables

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SEBA Class 9 Maths Chapter 4 Linear Equations in Two Variables

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Linear Equations in Two Variables

Chapter: 4

Exercise 4.1

Q.1. The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement. (Take the cost of a notebook to be Rs x and that of a pen to be Rs y)

Ans: Let the cost of a notebook is Rs. x and cost of a pen is Rs. y.

Therefore a linear equation in two variables (x and y) representing the given statement is x = 2y.

Q.2. Express the following linear equations in the form a x + by + c = 0 and indicate the values of a, b and c in each case.

Ans: 

On comparing the coefficients of x and y and also constant terms we observe that 

a=2,b=3 and c = – 

Ans: 

(iii) -2x+3y= 6

Ans: -2x+3y =6 in the form of

ax + by +c=0 is written as

-2x+3y-6=0 [Changing 6 to L.H.S.]

⇒ -2x+3y+(-6)=0

On comparing the coefficients of x and y and also constant terms we observe that

a=-2, b=3 and c=0

(iv) x = 3y

Ans: x=3y in the form of ax + by + c= 0 is written as

x-3y = 0 ⇒1.x+(-3)y-+0=0

On comparing the coefficients of x and y and also constant terms

we observe that a=1,b=-3 and c=0

(v) 2x = -5y

Ans: 2x=-5y in the form of ax + by + c= 0 is written

as 2x+5y =0 [Changing -5y to L.H.S.]

⇒ 2x+5y+0=0

On comparing the coefficients of x and y and also constant terms

we observe that a =2, b=5 and c= 0

(vi) 3x + 2 = 0

Ans: 3x + 2 = 0 in the form of ax + by + c = 0 is written as

3x+0.y +2 = 0

On comparing the coefficients of x and y and also constant terms

we observe that a =3, b=0 and c= 0

(vii) y – 2 = 0

Ans: y -2 = 0 in the form of ax + by + c = 0 is written as

0.x+1.y+(-2)=0

On comparing the coefficients of x and y and also constant terms

we observe that a =0, b=1 and c= -2

(viii) 5 = 2x

Ans: 5= 2x in the form of ax + by + c = 0 is written as 5-2.x=0

⇒ -2x+0.y+5=0

On comparing the coefficients of x and y and also constant terms

we observe that a =2, b = 0 and c=5

3. A linear equation in two variables x and y is of the form ax+by+c=0, where

(a) a =0,b =0

(b) a = 0,b = 0

(c) a = 0,b = 0

(d) a = 0,c = 0

Ans: . (a) a = 0,b = 0

Explanation :

If a ≠ 0,b ≠ 0, then the equation will be in the form of by+c=0 (which is form of one variable)

If a ≠ 0,b ≠ 0, then it will be in the form of ax+c=0 (which is form of one variable)

Exercise 4.2

Q.1. Which one of the following statements is true and why?

y = 3x + 5 has

(i) a unique solution.

Ans: We are given the equation y=3x+5 Put x = 0, then y=5

Therefore, (0,5) is the solution of this equation. Again, put x=1 then y =8

(ii) Only two solutions.

Ans: Therefore, (1, 8) is also the solution 

of this equation. Again, put x = 2 

then y=11.

(iii) Infinitely many solutions.

Ans: Therefore, (2, 11) is also the solution of this equation. Again x=3, y=8 and so on.

Now, it is clear that this equation has infinitely many solutions.

Remember. [A linear equation in two variable has infinitely many solution.]

Q.2. Write four solutions for each of the following equations:

(i) 2x + y = 7 

Ans: (i) We have 2x + y = 7 put x = 0, then y = 7

Therefore, (0, 7) is the solution of this equation.

Again put x = 1, then x = 5

Therefore (1, 5) is also the solution of this equation. Again put x= 2, 

then y = 3.

Therefore (2, 3) is also the solution of this equation. Again put x = 3

then y = 1

Therefore (3, 1) is also the solution of this equation. 

Therefore, (0,7), (1,5), (2, 3) and (3, 1) are

the four solutions of the equation 2x + y = 7

(ii) лx + y = 9

Ans: We have given πx+=9 or,

Therefore (0,9) is the solution of this equation.

Again put x = 7 they y = – 13

Therefore (7,-13) is also the solution of this equation. 

Again put x = 14, then y = – 35

Therefore (14,-35) is also the solution of this equation.

of equation πx+y=9

(iii) x+4y

Ans: We are given that x = 4y put y=0 then x=0

Therefore, (0, 0) is the solution of this equation.

Again y =1, then x=4.

Therefore (4, 1) is also part of this equation.

Again, put y = 2 then x= 8

Therefore, (8, 2) is also the solution of this equation.

Again put y = 3, x = 12

Therefore, (12, 3) is also the solution of this equation.

Hence, (0, 0), (4, 1), (8, 2) and (12, 3) are the four solutions of equation

x=4y

Q.3. Check which of the following are solutions of the equation

x-2y=4 and which are not.

(i) (0, 2) 

Ans: We are given that x-2y=4

put x=0 and y = 2

then, 0-2×2=4,⇒-4=4

Here, L.H.S.≠R.H.S

Therefore, (0, 2) is not the solution of equation x-2y=4

(ii) (2, 0) 

Ans: Put x=2 and y = 0

Then, 2-2×0=4⇒2=4

Here, L.H.S.≠R.H.S.

Therefore (2, 0) is not the solution of equation x-2y=4

(iii) (4,0) 

Ans: Put x=2 and y = 0

Then 4-2×0=4⇒4=4

Here, L.H.S.=R.H.S.

Therefore, (4, 0) is the solution of equation x-2y=4

(iv) (√2, 4√2) 

Ans: Put x=√2 and y = 4√2 in equation x-2y=4

Then √2-2×4√2=4 ⇒√2-8√2=4⇒7√2=4

Here, L.H.S.≠R.H.S.

Therefore, (√2, 4√2) is not the solution of equation x-2y=4

(v) (1, 1)

Ans: Put x=1 and y = 1 in equation x-2y=4

Then 1-2×1=4⇒1-2=4

L.H.S.≠R.H.S

Therefore, (1, 1) is not the solution of equation x – 2y = 4

2x + 2y = k.

Ans: We are given that x=2 and y = 1 is the solution of the equation

2x+3y=k

Put x=2 and y = 1 in equation 2x + 3y=k

2×2+3×1=k

4+3=k∴k=7

Hence, the value of k in equation 2x + 3y = k is 7

4. Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.

Ans: Putting x = 2 and y = 1 in equation 2x + 3y = k, we get

2.2 + 3.1 = k

   ⇒ 4 + 3 = k

   ⇒ 7 = k

Hence, the required value of k is 7.

5. If x = 3k + 2 and y = 2k − 1 is a solution of the equation 4x − 3y + 1 = 0. Find

the value of k.

Ans: Here, x = 3k + 2, y = 2k − 1

Now, 4x − 3y + 1 = 0

   ⇒ 4 (3k + 2) − 3 (2k − 1) + 1 = 0

   ⇒ 12 k + 8 − 6k + 3 + 1 = 0

   ⇒ 6 k + 12 = 0

   ⇒ 6k = − 12

   ⇒ k = − 2

∴ The value of k is − 2

6. For what value of k, x = 2k − 1 and y = k is a solution of the equation 3x − 5y − 7 = 0?

Ans: Now, 3x − 5y − 7 = 0

using the values of x and y we get

   3 (2k − 1) − 5k − 7 = 0

   ⇒ 6k − 3 − 5k − 7 = 0

   ⇒ k − 10 = 0

   ⇒ k = 10

∴ The value of k is 10

Choose the correct option (Q. No. 7 to 9)

7. Each of the points (− 2, 2), (0, 0), (3, −3) satisfies the linear equation

(a)   x − y = 0      

(b)   x + y = 0

(c) − x + 2y = 0    

(d)   x − 3y = 0

Ans. Correct option (b)

Reason :

When (− 2, 2), x + y = − 2 + 2 = 0

When (0, 0), x + y = 0 + 0 = 0

When (3, − 3), x + y = 3 + (− 3) = 0

8. Statement P : If (2, 0) is a solution of the linear equation 2x + 3y = k then the value of k is 4.

Statement Q : If (3, 4) is a solution of the linear equation 3y = ax + 7 then the value of a is 1/2.

(a) Both the statements P and Q are true

(b) Both the statements P and Q are false

(c) Statements P is true but Q is false

(d) Statements P is false but Q is true

Ans. Statement P :

      2x + 3y = K

Replacing x by 2 and y by 0 we get,

      2 × 2 + 3 × 0 = K

      ⇒ K = 4

Which is True.

Statement Q :

Here, 

3y = ax + 74. Find the value of k, if x = 2, y = 1 is a

solution of the equation 2x + 3y = k.

Sol. Putting x = 2 and y = 1 in equation 2x + 3y= k, we get

      2.2 + 3.1 = k

   ⇒ 4 + 3 = k

   ⇒ 7 = k

Hence, the required value of k is 7.

Replacing x by 3 and y by 4 we get 3 * 4 = a * 3 + 7 Statement Q is False Option (c) is correct Rightarrow 12 – 7 = 3a Rightarrow 5 = 3a a = 5/3

9. Consider the linear equation 2x – 3y = 12

(i) It has infinite number of solutions.

(ii) (5, 7) is one a solution of the equation.

(iii) (6,0) is not a solution of the equation.

(iv) (9, 2) is one of the solutions.

Choose the correct option:

(a) (i) and (iv)

(b) (ii) and (iv)

(c) (iii) and (iv)

(d) (i) and (iii)

Ans: Option (a) correct

Reason:

2x – 3y = 12 is a linear equation

Hence, it has infinite number of solutions.

option (i) is true.

When (9, 2), Replace x by 9 and y by 2 in the equation of L.H.S. 

= 2 * 9 – 3 * 2 

= 18 – 6 

= 12 =R.H.S

∴ option (iv) is true