# Solve system x^2-y=5 ; x - `sqrt y` = 1

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The given equations are: `x^2-y=5` --- (i)

and; `x -sqrty = 1` --- (ii)

From eqn.(ii), we get, `x=(1+sqrty)`

Substitute this value of x in eqn.(i)

`(1+sqrty)^2-y=5`

`rArr 1+2sqrty+y-y=5`

`rArr 2sqrty=5-1=4`

`rArr sqrty=2`

squaring both sides,

`y=4` .

Plugging in this value of y in eqn.(i),

`x^2-4=5`

`rArr x^2=(5+4)=9`

`rArr x=+-3`

Check:

Putting `x=3` , `y=4` , in eqn.(i)

`L.H.S. = 3^2-4=9-4=5=R.H.S.`

Putting `x=3` , `y=4` , in eqn.(ii)

`L.H.S. = 3-sqrt4=3-2=1=R.H.S.`

Both the equations hold good for this solution set.

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Putting` x=-3` , `y=4` , in eqn.(i)

`L.H.S. = (-3)^2-4=9-4=5=R.H.S.`

Putting `x=-3` , `y=4` , in eqn.(ii)

`L.H.S. = -3-sqrt4=-3-2=-5!=R.H.S.`

This is an extraneous solution to this system, hence discarded.

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**Therefore, the solution for the given system of equations is x=3, y=4.**

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