# NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2

These NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Exercise 4.2

Question 1.
Which one of the following statements is true and why?
y = 3x + 5 has
(i) a unique solution.
(ii) only two solutions.
(iii) infinitely many solutions.
Solution:
We have 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.
Therefore, (1, 8) is also the solution to this equation. Again, put x = 2, then y = 11.
Therefore, (2, 11) is also the solution of this equation. Again x = 3, x = 4 and so on.
Now, it is clear that this equation has infinitely many solutions.
Note: A linear equation in two variables has infinitely many solutions.

Question 2.
Write four solutions for each of the following equations.
(i) 2x + y = 7
(ii) πx + y = 9
(iii) x = 4y
Solution:
(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 y = 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) We have given πx + y = 9
or, $$\frac {22}{7}$$ x + y = 9 (∵ π = $$\frac {22}{7}$$)
or, y = 9 – $$\frac {22}{7}$$ x
Put x = 0 then y = 9
Therefore (0, 9) is the solution of this equation.
Again, put x = 1 then y = $$\frac {41}{7}$$
Therefore ($$\frac {22}{7}$$) is also the solution of this equation.
Again put x = 7, then y = -13
Therefore (7, – 13) is also the solution of this equation.
Again put x = 14, then y = -3
Therefore (14, -35) is also the solution of this equation.
Hence (0, 9) ($$\frac {22}{7}$$) (7, 13) and (14, -35) are the four solutions of equation πx + y = 9

(iii) We have given that x = 4y
Put y = 0, then x = 0
Therefore, (0, 0) is the solution of this equation.
Again put y = 1, then x = 4 Therefore, (4, 1) is also the solution of this equation.
Again, put y = 2 then x = 8 Therefore, (8, 2) is also the solution of this equation.
Again put y = 3, then 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 solution of equation x = 4y.

Question 3.
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)
(iv) (√2, 4√2)
(v) (1, 1)
Solution:
(i) we have 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) 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) Put x = 4 and y = 0 in equation x – 2y = 4
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) 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)
Put x = 1 and y = 1 in equation x – 2y = 4
Then 1 – 2 × 1 = 4
1 – 2 = 4
-1 = 4
L.H.S. ≠ R.H.S.
Therefore (1, 1) is not the solution of equation x – 2y = 4.

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

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