1. ## Simultaneous equations

x+y=7, x^2 - y^2 = 21
How do I do this?

2. Originally Posted by Detanon
x+7=7, xsquared - ysquared = 21
How do I do this?
Did you make a typo? I ask because I only see one equation with y in.

As is stands $\displaystyle x =0$ and $\displaystyle y = \pm i\sqrt{21}$

3. Do you not "see" that x=0 ?

4. woops sorry, I fixed it now.

5. Originally Posted by Detanon
woops sorry, I fixed it now.
As that's better

You can use the difference of two squares on the quadratic

$\displaystyle (x+y)(x-y) = 7(x-y) = 21 \: \:$

$\displaystyle \therefore \: \: x-y = 3 \: \: \rightarrow \: \: x = 3+y$

Sub in x=3+y into the first equation

(3+y)+y=7

Solve that for y then use the first equation (in either form) to find x

6. I dont get how you get (x+y)(x-y).

7. Originally Posted by Detanon
I dont get how you get (x+y)(x-y).
It is the difference of two squares.
Difference of two squares - Wikipedia, the free encyclopedia

From the difference of two squares we see that $\displaystyle x^2-y^2=(x+y)(x-y)$

If you expand the right hand side using FOIL the LHS is obtained: $\displaystyle (x+y)(x-y) = x^2-xy+xy-y^2$

Since $\displaystyle -xy+xy=0$ we get $\displaystyle (x+y)(x-y)=x^2-y^2$

============================

EDIT: If you prefer you can use the substitution method

$\displaystyle x = 7-y$

$\displaystyle (7-y)^2-y^2=21$

and solve that linear equation (y^2 cancels) but I think using the difference of two squares is easier