Saturday, May 30, 2020

NCERT solution class 9 chapter 4 Linear Equations in Two Variables exercise 4.2 mathematics

EXERCISE 4.2



Q 1, Ex 4.2, Page No 70, Linear Equations In Two Variables, CBSE Class 9th Mathematics

Page No 70:

Question 1:

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

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).


Q 2, Ex 4.2, Page No 70, Linear Equations In Two Variables, Maths Class 9th

Question 2:

Write four solutions for each of the following equations:
(i) 2x + y = 7
(ii) πx + y = 9
(iii) x = 4y

Answer:

(i) 2x + y = 7
For x = 0,
2(0) + = 7
⇒ = 7
Therefore, (0, 7) is a solution of this equation.
For = 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 = 2,
2(2) + y = 7
y = 3
Therefore, (2, 3) is a solution of this equation.

(ii) πx + y = 9
For x = 0,
π(0) + = 9
⇒ = 9
Therefore, (0, 9) is a solution of this equation.
For x = 1,
π(1) + = 9
= 9 − π
Therefore, (1, 9 − π) is a solution of this equation.
For x = 2,
π(2) + = 9
⇒ = 9 − 2π
Therefore, (2, 9 −2π) is a solution of this equation.
For x = −1,
π(−1) + = 9
⇒ = 9 + π
⇒ (−1, 9 + π) is a solution of this equation.

(iii) x = 4y
For x = 0,
0 = 4y
⇒ = 0
Therefore, (0, 0) is a solution of this equation.
For = 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.
For x = 2,
2 = 4y
⇒$y=\frac{2}{4}=\frac{1}{2}$
Therefore, $\left(2, \frac{1}{2}\right)$ is a solution of this equation.


Q 3, Ex 4.2, Page No 70, Linear Equations In Two Variables, CBSE Class 9th

Question 3:

Check which of the following are solutions of the equation x − 2= 4 and which are not:
(i) (0, 2)
(ii) (2, 0)
(iii) (4, 0)
(iv) $(\sqrt{2}, 4 \sqrt{2})$
(v) (1, 1)

Answer:

(i) (0, 2)
Putting x = 0 and y = 2 in the L.H.S of the given equation,
− 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,
− 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,
− 2= 4 − 2(0)
= 4 = R.H.S
Therefore, (4, 0) is a solution of this equation.

(iv) $(\sqrt{2}, 4 \sqrt{2})$
Putting
$x=\sqrt{2}$ and $y=4 \sqrt{2}$in the L.H.S of the given equation,
$x-2 y=\sqrt{2}-2(4 \sqrt{2})$
$=\sqrt{2}-8 \sqrt{2}=-7 \sqrt{2} \neq 4$

L.H.S ≠ R.H.S
Therefore, $(\sqrt{2}, 4 \sqrt{2})$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,
− 2y = 1 − 2(1) = 1 − 2 = − 1 ≠ 4
L.H.S ≠ R.H.S
Therefore, (1, 1) is not a solution of this equation.


Q 4, Ex 4.2, Page No 70, Linear Equations In Two Variables, Class 9th Maths

Question 4:

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

Answer:

Putting x = 2 and y = 1 in the given equation,
2+ 3y = k
⇒ 2(2) + 3(1) = k
⇒ 4 + 3 = k
⇒ k = 7
Therefore, the value of k is 7.

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