Question

√2x-√3y=1;√3x-√8y=0 by substitution method

Answer 1

Answer:

x = $\frac{\sqrt{8} }{7}$

y = $\frac{\sqrt{3} }{7}$

Step-by-step explanation:

Given,

√2x - √3y = 1    [ Let this be Eqn 1 ]

√3x - √8y = 0   [ Let this be Eqn 2 ]

From Eqn 2,

√3x = √8y

y = $\frac{\sqrt{3}x }{\sqrt{8} }$  [ Let this be Eqn 3 ]

Substituitng value of y in Eqn 1,

√2x - √3 ( $\frac{\sqrt{3}x }{\sqrt{8} }$ ) = 1

√2x + $\frac{3x}{\sqrt{8} }$ = 1

$\frac{\sqrt{2}x*\sqrt{8}+3x }{\sqrt{8} }$ = 1

√16 x + 3x = √8

4x + 3x = √8

7x = √8

Dividing both sides by 7,

7x / 7 = √8 / 7

x = $\frac{\sqrt{8} }{7}$

Squaring Eqn 3 we get,

( y = $\frac{\sqrt{3}x }{\sqrt{8} }$ )²

y² = $\frac{3x^{2}}{8}$

Substituting value of x in this Eqn,

y² = $\frac{3*(\frac{\sqrt{8} }{7} )^{2} }{8}$

y² = $\frac{3*8}{8*49}$

y² = $\frac{3}{49}$

y = $\sqrt{\frac{3}{49} }$

y = $\frac{\sqrt{3} }{7}$

Hence,

x = $\frac{\sqrt{8} }{7}$

y = $\frac{\sqrt{3} }{7}$