# Solve the simultaneous equation:

x+2y=3
x^2-4y^2=-33

## The Attempt at a Solution

x-2y=root-33?

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HallsofIvy
Homework Helper
First, that's not a solution, you haven't found a value for x and y! Second, sqrt(a- B) is NOT sqrt(A)- sqrt(B).

Probably it is simplest to solve the first equation for x, replace the "x" in the second equation by that expression.

Try this:
x+2y=3
=> x=3-2y

<< detailed solution hints deleted by berkeman >>

Last edited by a moderator:
LOL dude, use a^2 - b^2 = (a+b)(a-b)
and u have the a+b eqn. with you..
Get the a-b eqn from the above equation, and solve a+b and a-b for a and b.

HallsofIvy
Homework Helper
Oh, that's cute.

Mentallic
Homework Helper

It is always easiest (imo) to make x or y the subject in the linear equation: i.e. the first equation becomes $$x=3-2y$$ or $$y=\frac{3-x}{2}$$ and then substitute either of these values into the quadratic equation so that becomes: $$(3-2y)^2-4y^2=-33$$ or if you made y the subject: $$x^2-4(\frac{3-x}{2})^2=-33$$

Oh and remember you can't take the square root of a negative value.

x+2y=3
x^2-4y^2=-33
-33 = (x+2y)(x-2y)
got it now?

Mentallic
Homework Helper
Ahh the 4y2 slipped my mind. Didn't think difference of 2 squares would've had a factor equal to the linear equation.

I know, I wouldn't have found it until I was halfway through the problem. It was a nice catch.