Students can Download Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 9 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka State Syllabus Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.2

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.

Solution:

y = 3x + 5 has many solutions. Because this is a linear equation with two variables.

Question 2.

Write four solutions for each of the following equations :

i) 2x + y = 7

ii) πx + y = 9

iii) x = 4y

Solution:

i) 2x + y = 7

If x = 0, 2x + y = 7

2 × 0 + y = 7

0 + y = 7

y = 7

Solution is (0, 7).

If x = 1, 2x + y = 7

2 × 1 + y = 7

2 + y = 7

y = 7 – 2 = 5

Solution is (1, 5).

If x = 2, 2x + y = 7

2 × 2 + y = 7

4 + y = 7

y = 7 – 4 = 3

Solution is (2, 3).

If x = 3, 2x + y = 7

2 × 3 + y = 7

6 + y = 7

∴ y = 7 – 6 = 1

Solution is (3, 1).

ii) πx + y = 9

If x = 0, πx + y = 9

π × 0 + y = 9

0 + y = 9

y = 9

Solution (x, y) = (0, 0).

If x = 1, πx + y = 9

π × 1 + y = 9

π + y = 9

y = 9 – π

Solution (x, y) = (1, 9 – π).

If y = 0, πx + y = 9

πx + 0 = 9

πx = 9

∴ x = \(\frac{9}{\pi}\)

Solution (x, y) = ( \(\frac{9}{\pi}\), o).

If y = 1, πx + y = 9

πx + 1 = 9

πx = 9 – 1

πx = 8

∴ x = \(\frac{8}{\pi}\)

Solution (x, y) = (\(\frac{8}{\pi}\) ,1).

iii) x = 4y

If x = 0, x = 4y

0 = 4y

4y = 0

∴ y = \(\frac{0}{4}\) =∞

∴ Solution (x, y) = (0, ∞).

If x = 1, x = 4y

1 = 4y

4y = 1

∴ y = \(\frac{1}{4}\)

∴ Solution (x, y) = (1, \(\frac{1}{4}\)).

If x = 2, x = 4y

2 = 4y

4y = 2

∴ y = \(\frac{4}{2}\)

∴ y = \(\frac{1}{2}\)

Solution (x, y) = (2, \(\frac{1}{2}\)).

If x = 4, x = 4y

4 = 4y

4y = 4

∴ y = \(\frac{4}{4}\)

∴ y = 1

∴ Solution (x, y) = (4, 1).

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) (\(\sqrt{2}, 4 \sqrt{2}\))

v) (1, 1)

Solution:

i) (0, 2)

x – 2y = 4

0 – 2(2) = 4

0 – 4 = 4

-4 ≠ 4

∴ (0, 2) is not a solution.

ii) (2, 0)

x – 2y = 4

2 – 2(0) = 4

2 – 0 = 4

2 ≠ 4

∴ (2, 0) is not a solution

iii) (4, 0)

x – 2y = 4

4 – 2(0) = 4

4 – 0 = 4

4 = 4

∴ (4, 0) is a solution.

iv) Put x = √2, y = 4√2 in equation (i)

x – 2y = 4

√2 – 2 (4√2) = 4

√2 – 8√2 = 4

-7√2 = 4

but -7√2 ≠ 4

∴ (√2, 4√2) is not the solution of given equation.

v) (1, 1)

x – 2y = 4

1 – 2(1) = 4

1 – 2 = 4

-1 ≠ 4

∴ (1, 1) is not a solution.

Question 4.

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

Solution:

2x + 3y = k

If x = 2, y = 1, then k = ?

2x + 3y = k

2(2) + 3(1) = k

4 + 3 = k

7 = k

∴ k = 7.