Tamil Nadu Board of Secondary EducationSSLC (English Medium) Class 9th

# Solve, using the method of substitution 2x-3y = 1, 3x-8y = 0 - Mathematics

Sum

Solve, using the method of substitution

sqrt(2)x - sqrt(3)y = 1, sqrt(3)x - sqrt(8)y = 0

#### Solution

sqrt(2)x - sqrt(3)y = 1

- sqrt(3)y = 1 - sqrt(2)x

sqrt(3)y = sqrt(2)x - 1

y = (sqrt(2)x - 1)/sqrt(3) → (1)

sqrt(3)x - sqrt(8)y = 0 → (2)

Substitute the value of y in (2)

sqrt(3)x - (sqrt(8)(sqrt(2)x - 1))/sqrt(3)

multiply by sqrt(3)

⇒ (3x - sqrt(8)(sqrt(2)x - 1))/sqrt(3)

3x - sqrt(8)(sqrt(2)x - 1) = 0

3x - 4x + sqrt(8) = 0

− x = sqrt(8)

Substitute the value of x in (1)

y = (sqrt(2) xx sqrt(8) - 1)/sqrt(3)

= (4 - 1)/sqrt(3)

= 3/sqrt(3)

= sqrt(3)((3 xx sqrt(3))/(sqrt(3) xx sqrt(3)))

The value of x = sqrt(8) and y = sqrt(3)

Concept: Methods of Solving Simultaneous Linear Equations by Substitution
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#### APPEARS IN

Samacheer Kalvi Mathematics Class 9th Tamil Nadu State Board
Chapter 3 Algebra
Exercise 3.11 | Q 1. (iv) | Page 126