SOLUTION: I'm not sure if this is the right section, but will someone please help me. Find the exact solutions of the system of equations x^2+2y^2=18 and x=2y. I have to show all my work a

Question 167397: I'm not sure if this is the right section, but will someone please help me.
Find the exact solutions of the system of equations x^2+2y^2=18 and x=2y.
I have to show all my work and I don't know how to do this.
PLEASE HELP!!!!
Thanks
Found 2 solutions by Fombitz, gonzo:
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
1.
2.
Substitute eq. 2 into eq. 1, directly substitute for x and solve for y.
Then go back and find x.
1.

Then from eq. 2,
2.

.
.
.
When we graph the equations, you can see the solutions more clearly.
Equation 1 is the ellipse.
Equation 2 is the straight line.
.
.
.
(1.73,3.46) and (-1.73,-3.46)
.
.
.

Answer by gonzo(654) (Show Source): You can put this solution on YOUR website!
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)
-----
if y = +/- sqrt(3), then x = 2y = +/- 2*sqrt(3)
-----
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.
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