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Linear Equations in Two Variables

*Exercise 4.2***Q1: Which one of the following options is true, and why?**

y = 3x + 5 has

(i) a unique solution, (ii) only two solutions, (iii) infinitely many solutions

y = 3x + 5 has

(i) a unique solution, (ii) only two solutions, (iii) infinitely many solutions

Answer:

y = 3x + 5 is a linear equation in two variables and it has infinite possible solutions. As for every value of

x, there will be a value of y satisfying the above equation and vice-versa.

Hence, the correct answer is (iii).

**Q2: Write four solutions for each of the following equations:**

(i) 2x + y = 7 (ii) ∏x + y = 9 (iii) x = 4y

(i) 2x + y = 7 (ii) ∏x + y = 9 (iii) x = 4y

Answer:

**(i) 2x + y = 7**

For x = 0,

2(0) + y = 7

⇒ y = 7

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

For x = 1,

2(1) + y = 7

⇒ y = 5

Therefore, (1, 5) is a solution of this equation.

For x = −1,

2(−1) + y = 7

⇒ y = 9

Therefore, (−1, 9) is a solution of this equation.

For x = 2,

2(2) + y = 7

⇒ y = 3

Therefore, (2, 3) is a solution of this equation.

**(ii) ∏x + y = 9**

For x = 0,

∏(0) + y = 9

⇒ y = 9

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

For x = 1,

∏(1) + y = 9

⇒y = 9 − ∏

Therefore, (1, 9 − ∏) is a solution of this equation.

For x = 2,

∏(2) + y = 9

⇒ y = 9 − 2∏

Therefore, (2, 9 −2∏) is a solution of this equation.

For x = −1,

∏(−1) + y = 9

⇒ y = 9 + ∏

⇒ (−1, 9 + ∏) is a solution of this equation.

0 = 4y

⇒ y = 0

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

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

For x = 1,

∏(1) + y = 9

⇒y = 9 − ∏

Therefore, (1, 9 − ∏) is a solution of this equation.

For x = 2,

∏(2) + y = 9

⇒ y = 9 − 2∏

Therefore, (2, 9 −2∏) is a solution of this equation.

For x = −1,

∏(−1) + y = 9

⇒ y = 9 + ∏

⇒ (−1, 9 + ∏) is a solution of this equation.

**(iii) x = 4y**For x = 0,0 = 4y

⇒ y = 0

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

For y = 1,

x = 4(1) = 4

Therefore, (4, 1) is a solution of this equation.

For y = −1,

x = 4(−1)

⇒ x = −4

Therefore, (−4, −1) is a solution of this equation.

x = 4(1) = 4

Therefore, (4, 1) is a solution of this equation.

For y = −1,

x = 4(−1)

⇒ x = −4

Therefore, (−4, −1) is a solution of this equation.

For x = 2,

2 = 4y

y = 2/4 = 1/2

Therefore,(2 , 1/2) is a solution of this equation.

Answer:

(i) (0, 2)

Putting x = 0 and y = 2 in the L.H.S of the given equation,

x − 2y = 0 − 2 ☓ 2 = − 4 ≠ 4

L.H.S ≠ R.H.S

Therefore, (0, 2) is not a solution of this equation.

(ii) (2, 0)

Putting x = 2 and y = 0 in the L.H.S of the given equation,

x − 2y = 2 − 2 ☓ 0 = 2 ≠ 4

L.H.S ≠ R.H.S

Therefore, (2, 0) is not a solution of this equation.

(iii) (4, 0)

Putting x = 4 and y = 0 in the L.H.S of the given equation,

x − 2y = 4 − 2(0) = 4 = R.H.S

Therefore, (4, 0) is a solution of this equation.

(iv) (√ 2 , 4√ 2 )

2 = 4y

y = 2/4 = 1/2

Therefore,(2 , 1/2) is a solution of this equation.

**Q3: Check which of the following are solutions of the equation x − 2y = 4 and which are not:**

(i) (0, 2 (ii) (2, 0) (iii) (4, 0)(i) (0, 2 (ii) (2, 0) (iii) (4, 0)

**(iv) (√ 2 , 4√ 2 ) (v)(1, 1)**Answer:

(i) (0, 2)

Putting x = 0 and y = 2 in the L.H.S of the given equation,

x − 2y = 0 − 2 ☓ 2 = − 4 ≠ 4

L.H.S ≠ R.H.S

Therefore, (0, 2) is not a solution of this equation.

(ii) (2, 0)

Putting x = 2 and y = 0 in the L.H.S of the given equation,

x − 2y = 2 − 2 ☓ 0 = 2 ≠ 4

L.H.S ≠ R.H.S

Therefore, (2, 0) is not a solution of this equation.

(iii) (4, 0)

Putting x = 4 and y = 0 in the L.H.S of the given equation,

x − 2y = 4 − 2(0) = 4 = R.H.S

Therefore, (4, 0) is a solution of this equation.

(iv) (√ 2 , 4√ 2 )

Putting x = √ 2 and y = 4√ 2 in the L.H.S of the given equation,

x - 2y = √ 2 - 2(4√ 2 )

= √ 2 - 8√ 2 = -7√ 2 ≠ 4

L.H.S ≠ R.H.S

Therefore,(√ 2 , 4√ 2 ) is not a solution of this equation.

(v) (1, 1)

Putting x = 1 and y = 1 in the L.H.S of the given equation,

x − 2y = 1 − 2(1) = 1 − 2 = − 1 ≠ 4

L.H.S ≠ R.H.S

Therefore, (1, 1) is not a solution of this equation.

x - 2y = √ 2 - 2(4√ 2 )

= √ 2 - 8√ 2 = -7√ 2 ≠ 4

L.H.S ≠ R.H.S

Therefore,(√ 2 , 4√ 2 ) is not a solution of this equation.

(v) (1, 1)

Putting x = 1 and y = 1 in the L.H.S of the given equation,

x − 2y = 1 − 2(1) = 1 − 2 = − 1 ≠ 4

L.H.S ≠ R.H.S

Therefore, (1, 1) is not a solution of this equation.

**Q4: Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.**
Answer:

Putting x = 2 and y = 1 in the given equation,

2x + 3y = k

⇒ 2(2) + 3(1) = k

⇒ 4 + 3 = k

⇒ k = 7

Therefore, the value of k is 7.

2x + 3y = k

⇒ 2(2) + 3(1) = k

⇒ 4 + 3 = k

⇒ k = 7

Therefore, the value of k is 7.

its good

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