NCERT Solutions Class 9 Maths Chapter 4 Linear Equations In Two Variable Exercise 4.2
Introduction:
In this exercise/article we will learn about Linear Equations In Two Variable. You have seen that every linear equation in one variable has a unique solution. What can you say about the solution of a linear equation involving two variables? As there are two variables in the equation, a solution means a pair of values, one for x and one for y which satisfy the given equation.
NCERT Class 9 Maths Chapter 4 Coordinate Geometry :
Class 9 Maths Exercise 4.2 (Page-59)
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
Solution :
(iii) infinitely many solutions
∴ It has infinitely many solutions because for same value of x there will be some value of y will exist .
Q2. Write four solutions for each of the following equations:
(i) 2x + y = 7
Solution :
We have 2x + y = 7
According to the question,
putting value of x = 0
Now, (i) 2 × 0 + y = 7
⇒ 0 + y = 7
⇒ y = 7
(ii) putting value of x = 1
⇒ 2 × 1 + y = 7
⇒ 2 + y = 7
⇒ y = 7 - 2
⇒ y = 5
(iii) putting value of x = 2
⇒ 2 × 2 + y = 7
⇒ 4 + y = 7
⇒ y = 7 - 4
⇒ y = 3
(iv) putting value of x = 3
⇒ 2 × 3 + y = 7
⇒ 6 + y = 7
⇒ y = 7 - 6
⇒ y = 1
∴ ( 0, 7 ) , ( 1, 5 ) , ( 2, 3 ) and ( 3, 1 ) are the four solutions of the equation 2x + y = 7 .
(ii) πx + y = 9
Solution :
We have πx + y = 9
According to the question,
putting value of x = 0
Now, (i) πx + y = 9
⇒ π × 0 + y = 9
⇒ 0 + y = 9
⇒ y = 9
(ii) putting value of x = 1
⇒ π × 1 + y = 9
⇒ π + y = 9
⇒ y = 9 - π
(iii) putting value of x = 2
⇒ π × 2 + y = 9
⇒ 2π + y = 9
⇒ y = 9 - 2π
(iv) putting value of x = 3
⇒ π × 3 + y = 9
⇒ 3π + y = 9
⇒ y = 9 - 3π
∴ ( 0, 9 ) , ( 1, 9 - π ) , ( 2, 9 - 2π ) and ( 3, 9 - 3π ) are the four solutions of the equation πx + y = 9 .
(iii) x = 4y
Solution :
We have x = 4y
According to the question,
putting value of y = 0
Now, (i) x = 4y
⇒ x = 4 × 0
⇒ x = 0
(ii) putting value of y = 1
⇒ x = 4 × 1
⇒ x = 4
(iii) putting value of y = 2
⇒ x = 4 × 2
⇒ x = 8
(iv) putting value of y = 3
⇒ x = 4 × 3
⇒ x = 12
∴ ( 0, 0 ) , ( 4, 1 ) , ( 8, 2 ) and ( 12, 3 ) are the four solutions of the equation x = 4y .
Q3. Check which of the following are solutions of the equation x - 2y = 4 and which are not:
(i) ( 0, 2 )
Solution :
Given equation x - 2y = 4
Given values ⇒ x = 0, y = 2
According to the question,
Here, x - 2y = 4
put values of x and y
Now, 0 - 2 × 2 = 4
⇒ 0 - 4 = 4
⇒ - 4 = 4
LHS \(\displaystyle \ne \) RHS
∴ ( 0, 2 ) is a not solutions of the equation x - 2y = 4 .
(ii) ( 2, 0 )
Solution :
Given equation x - 2y = 4
Given values ⇒ x = 2, y = 0
According to the question,
Here, x - 2y = 4
put values of x and y
⇒ 2 - 2 × 0 = 4
⇒ 2 - 0 = 4
⇒ 2 = 4
LHS \(\displaystyle \ne \) RHS
∴ ( 2, 0 ) is a not solutions of the equation x - 2y = 4 .
(iii) ( 4, 0 )
Solution :
Given equation x - 2y = 4
Given values ⇒ x = 4, y = 0
According to the question,
Here, x - 2y = 4
put values of x and y
⇒ 4 - 2 × 0 = 4
⇒ 4 - 0 = 4
⇒ 4 = 4
LHS = RHS
∴ ( 4, 0 ) is a solutions of the equation x - 2y = 4 .
(iv) ( \(\displaystyle \sqrt{2}\), \(\displaystyle 4\sqrt{2}\) )
Solution :
Given equation x - 2y = 4
Given values ⇒ x = \(\displaystyle \sqrt{2}\), y = \(\displaystyle 4\sqrt{2}\)
According to the question,
Here, x - 2y = 4
put values of x and y
⇒ \(\displaystyle \sqrt{2}\) - 2 × \(\displaystyle 4\sqrt{2}\) = 4
⇒ \(\displaystyle \sqrt{2}\) - \(\displaystyle 8\sqrt{2}\) = 4
⇒ \(\displaystyle -7\sqrt{2}\) = 4
LHS \(\displaystyle \ne \) RHS
∴ ( \(\displaystyle \sqrt{2}\), \(\displaystyle 4\sqrt{2}\) ) is a not solutions of the equation x - 2y = 4 .
(v) ( 1, 1 )
Solution :
Given equation x - 2y = 4
Given values ⇒ x = 1, y = 1
According to the question,
Here, x - 2y = 4
put values of x and y
⇒ 1 - 2 × 1 = 4
⇒ 1 - 2 = 4
⇒ - 1 = 4
LHS \(\displaystyle \ne \) RHS
∴ ( 1, 1 ) is a not solutions of the equation x - 2y = 4 .
Q4. Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
Solution :
Given equation 2x + 3y = k
Given value ⇒ x = 2, y = 1
According to the question
Here, 2x + 3y = k
put values of x and y
Now, 2 × 2 + 3 × 1 = k
⇒ 4 + 3 = k
⇒ 7 = k
∴ The value of k = 7 .