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

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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
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