Question 1061541
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Please help me solve this by using the substitution method. Thank you! 

{{{ x^2+y^2=90 }}}

{{{ y=sqrt(x) }}}
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{{{x^2 + y^2}}} = 90,    (1)
y = {{{sqrt(x)}}}.         (2)

Square equation (2) (both sides). You will get

{{{y^2}}} = x.

Using this, replace {{{y^2}}} by "x" in the equation (2). You will get

{{{x^2 + x}}} = 90,      (1')   or

{{{x^2 + x - 90}}} = 0.

Solve by using the quadratic formula. You will get

{{{x[1,2]}}} = {{{(-1 +- sqrt (1 + 4*90))/2}}} = {{{(-1 +- 19)/2}}}.


Thus {{{x[1]}}} = 9,  {{{x[2]}}} = -10.


Since "x" under the square root must be non-negative, only the root x= 9 survives.

Then y = {{{sqrt(x)}}} = +/- 3.


<U>Answer</U>. There are 2 solutions: (x,y) = (9,3)  and  (x,y) = (9,-3).
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