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To solve the system
we need to substitute the second equation into the first, then solve the subsequent quadratic equation.
Upon substitution, we see that
`9(2y)+4(y-2)^2=36` now expand
`18y+4(y^2-4y+4)=36` expand again
`18y+4y^2-16y+16=36` collect like terms
`4y^2+2y-20=0` divide by 2
`2y^2+y-10=0` factor left side
`(2y+5)(y-2)=0` now solve to get:
`y=2` or `y=-5/2` .
Now we need to substitute the y-values into the original equations, but since `x^2=2y` , this means that `y>0` , so the only possible solution is `y=2` .
Now solving for x, we get
This means there are two solutions `x=2` , `y=2` and `x=-2` , `y=2` . As a check, note that `9(+-2)+4(2-2)^2=36` .
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