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

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$(a-b)^2\ne a^2-b^2$, though squaring seems a good place to start. And $a^2+a^2\ne a^4$. Does $3^2+3^2=3^4$? – Mike Apr 15 '13 at 6:45
Oops.. I don't know what I was thinking :S Thanks for corecting that. – math_is_fun Apr 16 '13 at 3:26
@math_is_fun Your question now doesn't make sense. Please put your full question back into the text. – Daryl Apr 16 '13 at 5:10
@Daryl Oh no. So sorry :\ – math_is_fun Apr 16 '13 at 5:31

Given:

$a - b = 3$ equation of a line

$a^2 + b^2 = 120$ equation of a circle.

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:

$b^2 + 6b+9+ b^2 = 120\, \Rightarrow 2b^2+6b+9 = 120$

$2b^2 + 6b -111 = 0$

$b = \left(1/2\right) \left(3 \pm \sqrt{231}\right)$

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:

1. $a - b = 3$
2. $a^2 + b^2 = 120$

from equation 1, $a = b + 3$

Subitute $a = b + 3$ into equation 2. Resulting in $\left(b + 3\right)^2 + b^2 = 120$

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.
I followed you up to "Use the quadratic equation". From here, I used the form $$ax^2 + bx + c = 0$$ and applied a, b, c to $$x = \frac{(-b ± sqrt(b^2 - 4ac))}{2a}$$ If I'm correct so far, I end up with $$x = (-6 ± \sqrt924)/4 ⇒ (\frac{-3}{2} ± \sqrt231)$$ I'm not sure how you got (1/2) or -1/3 in $$b=(\frac12)(\frac{-1}{3}±\sqrt231)$$ – math_is_fun Apr 16 '13 at 3:35
The fourth root was a silly mistake; @Mike pointed it out in his comment. I will edit my OP to show that my working was flawed and I'm only following your steps above to understand the solution. WRT to 924, I get this by taking a, b, c from the quadratic equation $2b^2 + 6b − 111 = 0$ and subtituting them in $\sqrt{b^2 - 4ac}{2a} ⇒ \sqrt{6^2 - 4(2)(-111)} ⇒ \sqrt{924}$ – math_is_fun Apr 16 '13 at 4:47
@math_is_fun Ok, just making sure $\sqrt{924} = 2\sqrt{231}$ – yiyi Apr 16 '13 at 4:49