Can this be solved as a simultaneous equation? I have a homework question for algebra class and I'm not sure how to solve it; I thought it was a simultaneous equation but after working through it, I'm not sure. I'd appreciate a push in the right direction.
For the equations below, I have to find the value of $a$ and $b$.
Equation 1: $a - b = 3$
Equation 2: $a^2 + b^2 = 120$
 A: Given: 
$a - b = 3$ equation of a line
$a^2 + b^2 = 120$ equation of a circle. 
well, here is your answer. 
Try it.  As this is being homework, I will post this teaser for you. 
let $a = b+3$ and then replace the $a$ of the 2nd equation and solve for $b$ and then...
$\left(b+3\right)^2 + b^2 = 120$ 
Expand: 
$b^2 + 6b+9+ b^2 = 120\, \Rightarrow 2b^2+6b+9 = 120$ 
Use the quadratic equation 
$2b^2 + 6b -111 = 0$
$b = \left(1/2\right) \left(3 \pm \sqrt{231}\right)$
plug in to the first equation and find that $a = \left(1/2\right) \left(3 \mp \sqrt{231}\right)$ 
note that $a$ and $b$ have oppsite signs before the square root. 
Here is a picture of your solution: 


I'll try to be more clear. 
You have two equations: 


*

*$a - b = 3$ 

*$a^2 + b^2 = 120$ 


from equation 1, $a = b + 3$ 
Subitute $a = b + 3$ into equation 2.  Resulting in $\left(b + 3\right)^2 + b^2 = 120$ 
Then follow my instruction above. 
I don't know why you think that $a^2 + b^2 = 9$?  That is not correct for the given equations 1 and 2. 
Your step that $a^4 = 129$ is incorrect and not valid. 
