NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.2
NCERT Solutions for Class 9 Maths Chapter 4 – Linear Equations in Two Variables Ex 4.2 are given here in simple, step-by-step form. NCERT stands for National Council of Educational Research and Training, the body that makes your textbook. A linear equation in two variables is an equation that can be written as ax + by + c = 0, where x and y are the two variables. Exercise 4.2 answers one main question: how many solutions does such an equation have?
This exercise has 4 questions. It teaches an important idea. A linear equation in one variable has only one solution, but a linear equation in two variables has infinitely many solutions. To find a solution, you choose any value for one variable, put it into the equation, and work out the other variable. A free printable PDF of these solutions is also available for offline study.
Exercise 4.2 – Questions and Answers
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
Answer:
Since y = 3x + 5 is a linear equation in two variables, it has infinite possible solutions. For every value of x, there will be a value of y satisfying the above equation, and vice-versa.
For example, if x = 0, then y = 3(0) + 5 = 5, so (0, 5) is a solution.
If x = 1, then y = 3(1) + 5 = 8, so (1, 8) is a solution.
If x = −2, then y = 3(−2) + 5 = −1, so (−2, −1) is a solution.
Since x can be any real number, the list never ends.
Therefore, the correct option is (iii) infinitely many solutions.
Q2. 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
or y = 7 − 2x
We can get 4 solutions of the equation by putting x = 0, 1, 2, 3
y = 7 − 2(0) = 7
y = 7 − 2(1) = 5
y = 7 − 2(2) = 3
y = 7 − 2(3) = 1
Therefore, the four solutions of the given equation are:
(0, 7), (1, 5), (2, 3) and (3, 1).
(ii) πx + y = 9
y = 9 − πx
We can get 4 solutions of the equation by putting x = 0, 1, 2, 3
y = 9 − π(0) = 9
y = 9 − π(1) = 9 − π
y = 9 − π(2) = 9 − 2π
y = 9 − π(3) = 9 − 3π
Therefore, the four solutions of the given equation are:
(0, 9), (1, 9 − π), (2, 9 − 2π) and (3, 9 − 3π).
(iii) x = 4y
We can get 4 solutions of the equation by putting y = 0, 1, 2, 3
x = 4(0) = 0
x = 4(1) = 4
x = 4(2) = 8
x = 4(3) = 12
Therefore, the four solutions of the given equation are:
(0, 0), (4, 1), (8, 2) and (12, 3).
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) (iv) (√2, 4√2) (v) (1, 1)
Answer:
(i) (0, 2)
Putting x = 0 and y = 2 in L.H.S. of x − 2y = 4, we get
L.H.S. = x − 2y = 0 − 2(2) = −4 ≠ R.H.S.
Therefore, (0, 2) is not a solution of the given equation.
(ii) (2, 0)
Putting x = 2 and y = 0 in L.H.S. of x − 2y = 4, we get
L.H.S. = x − 2y = 2 − 2(0) = 2 ≠ R.H.S.
Therefore, (2, 0) is not a solution of the given equation.
(iii) (4, 0)
Putting x = 4 and y = 0 in L.H.S. of x − 2y = 4, we get
L.H.S. = x − 2y = 4 − 2(0) = 4 = R.H.S.
Therefore, (4, 0) is a solution of the given equation.
(iv) (√2, 4√2)
Putting x = √2 and y = 4√2 in L.H.S. of x − 2y = 4, we get
L.H.S. = x − 2y
= √2 − 2(4√2)
= √2 − 8√2
= −7√2 ≠ R.H.S.
Therefore, (√2, 4√2) is not a solution of the given equation.
(v) (1, 1)
Putting x = 1 and y = 1 in L.H.S. of x − 2y = 4, we get
L.H.S. = x − 2y = 1 − 2(1) = −1 ≠ R.H.S.
Therefore, (1, 1) is not a solution of the given 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 2x + 3y = k, we get
2(2) + 3(1) = k
4 + 3 = k
Or k = 7
Therefore, the value of k is 7.