Question 167397
here's how i would do it.
original equation:
x^2+2y^2=18 and x=2y.
since x = 2y, equation becomes:
(2y)^2 + 2y^2 = 18
this becomes:
4y^2 = 2y^2 = 18
this becomes:
6y^2 = 18
divide both sides by 6 to get:
y^2 = 3
y = +/- sqrt(3)
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if y = +/- sqrt(3), then x = 2y = +/- 2*sqrt(3)
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to prove the answer is correct, substitute in original equation.
original equation is:
x^2+2y^2=18
substitute 2*sqrt(3) for x and sqrt(3) for y:
(2*sqrt(3))^2 + 2*(sqrt(3))^2 = 18
this becomes:
4*3 + 2*3 = 18
which becomes:
12 + 6 = 18
which becomes:
18 = 18 proving answer is correct.
it doesn't matter if + or - sqrt(3) is used since when you square it, it comes out positive either way.