Solving for three unknowns given three linear equations Have to admit, I forgot how to do simple 9th grade algebra problems. I need to know how to do this stuff for an exam tomorrow.  I am doing recurrences and the only thing I need to do is find $a, b$, and $c$ from the following
$$\begin{align*}
a + b + c &= 3\\
4a + 2b+ c&= 6\\
9a + 3b + c& = 13
\end{align*}$$
Can someone refresh my memory on how to solve for  $a, b$ and $c$
?

Just use Gaussian elimination.
$a+b+c=3$ 
$4a+2b+c=6$
$9a+3b+c=13$
Subtracting equation 1 from equation 2 and equation 3:
$a+b+c=3$ 
$3a+b+0c=3$
$8a+2b+0c=10$
Subtract 2 times equation 2 from equation 3.
$a+b+c=3$ 
$3a+b+0c=3$
$2a+0b+0c=4$
We can see that $a=2$
Substitute this into the other equations
$2+b+c=3$ 
$3(2)+b+0c=3$
$a=2$
We can now see $b=-3$ Substitute this into equation 1.
$2+(-3)+c=3$ 
$b=-3$
$a=2$
Which shows that $c=4$
Testing these by putting them into the original equations.
$2+-3+4=3$
$4(2)+2(-3)+4=6$
$9(2)+3(-3)+4=13$
And we see they solve the system.
