Finding coefficients of quadratic given one tangent and point on the curve

I am given a quadratic equation: $$y = Ax^2 + Bx + C$$ that passes through $(1,3)$ and $(2,3)$, and a tangent to the curve is $x - y + 1 = 0$ at $(2.3)$.

How do I find $A$, $B$, and $C$?

The derivative of $\mathrm dy/\mathrm dx = 2AX + B$, so at $x=2$, the slope of the tangent is $4A + B$, and from the givens we know that $4A + B = 1$. We also know that $$3 = A + B + C,\qquad 3 = 4A + 2B + C.$$

From there, how does one find $A$, $B$, and $C$?

(I can't seem to get the answers that make any sense from here).

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Putting, $4a+b=1$ in $4a+b+b+c=3$ gives $b+c=2$, putting this in $a+b+c=3$ gives $a=1$
Now, $b=1-4a=-3$ and $c=2-b=5$
General method to solve such equation can be found by searching linear equation in $3$ variables if you are familiar with linear algebra (matrices, determinants, etc).