# Solve the simultaneous equation:

1. Oct 23, 2008

### d3v3st4t10n

1. The problem statement, all variables and given/known data
x+2y=3
x^2-4y^2=-33

2. Relevant equations

3. The attempt at a solution
x-2y=root-33?

2. Oct 23, 2008

### HallsofIvy

Staff Emeritus
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.

3. Nov 20, 2008

### v0id19

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

<< detailed solution hints deleted by berkeman >>

Last edited by a moderator: Nov 20, 2008
4. Nov 20, 2008

### spideyunlimit

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.

5. Nov 21, 2008

### HallsofIvy

Staff Emeritus
Oh, that's cute.

6. Nov 21, 2008

### Mentallic

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.

7. Nov 21, 2008

### spideyunlimit

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

8. Nov 21, 2008

### Mentallic

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

9. Nov 21, 2008

### Chaos2009

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