Solve the following system of equations $$\large \left\{ \begin{align} (x + y)^2 + x^2y + 7x + 4 &= x^3 + 7y\\ 3x^2 + y^2 + 8y + 4 &= 8x \end{align}\right.$$
Here's what I did. (Sincerely, I feel very unintelligent inside.)
From the system of equations, we have that $$\left\{ \begin{align} x^3 - x^2y - x^2 - y^2 - 2xy - 8(x - y) + x - y - 4 &= 0\\ 3x^2 + y^2 + 4 = 8(x - y)\end{align} \right.$$
$\implies x^3 - x^2y - 4x^2 - 2y^2 - 2xy + x - y - 8 = 0$
$\iff x^2(x - y - 4) - (2y - 1)(x - y) - 4y^2 - 8 = 0$
And I gave up, already. Such laziness is coming out of this particular brain.
Hint: Computing equation one minus equation two and factorizing we get $$(x+5)(x-3)(x-y)=0$$ Can you proceed?