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Question
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)`
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