System of equations - what are the values of the coefficients of a quadratic parabola? In the process of relearning the mathematical basics I'm stumbling over this problem:

A quadratic parabola $y = ax^2 + bx + c $ goes through the points A(1/2), B(3/7) and C(-1/1). What are the values of the coefficients $a$, $b$ and $c$?

This is a problem presented in the section about "Systems of Equalities", but I don't have the slightest idea, how to use the coordinates to calculate the values of the coefficients.
How can I solve this problem with a system of equalities?

First use the fact that the curve passes through the point $(1,2)$. That says that when $x=1$, we have $y=2$. So substitute $x=1$ in the equation $y=ax^2+bx+c$. We get
$$2=a+b+c.$$
Similarly, because $(3,7)$ is on the curve, we have
$$7=9a+3b+c.$$
And finally, the third point tells us that 
$$1=a-b+c.$$
We now have $3$ linear equations in the $3$ unknowns $a$, $b$ and $c$. 
From the two equations $9a+3b+c=7$ and $a+b+c=2$, we obtain, by subtraction, that 
$8a+2b=5$.
From the two equations $9a+3b+c=7$ and $a-b+c=1$, we obtain, by subtraction, that $8a+4b=6$. (I deliberately did not subtract the third from the first, that would have made things too easy!)
We have "eliminated" $c$, and we have two equations in the variable $a$ and $b$. Now we will "eliminate" $b$, which again is not the clever thing to do.
So recall we have $8a+2b=5$ and $8a+4b=6$. Multiply both sides of the first equation by $2$. We get $16a+4b=10$. By subtraction, using $8a+4b=6$, we get $8a=4$, and therefore $a=1/2$. Then from $8a+2b=5$ we get $4+2b=5$ and therefore $b=1/2$. Finally, from $a+b+c=2$ we get that $c=1$.
