[Algebra]
Problem: Suppose that the functions $f(x)$ and $g(x)$ satisfy the system of equations

$$
\begin{aligned}
f(x)+3 g(x) & =x^{2}+x+6 \\
2 f(x)+4 g(x) & =2 x^{2}+4
\end{aligned}
$$

for all $x$. Determine the values of $x$ for which $f(x)=g(x)$.

Solution: We solve the system of equations for $f(x)$ and $g(x)$.

Dividing out the common factor of 2 from the second equation, we get

$f(x)+2 g(x)=x^{2}+2$.

Subtracting from the first equation, we get $g(x)=x+4$.

Thus, $f(x)=x^{2}+2-2 g(x)=x^{2}+2-2(x+4)=x^{2}-2 x-6$.

Equating $f(x)$ and $g(x)$, we obtain

$$
\begin{aligned}
x^{2}-2 x-6 & =x+4 \\
x^{2}-3 x-10 & =0 \\
(x-5)(x+2) & =0
\end{aligned}
$$

Therefore, $x=5$ or $x=-2$.
Instead of considering the equation $f(x)=g(x)$, we consider the equation $f(x)-g(x)=0$, and we try to obtain an expression for $f(x)-g(x)$ by manipulating the two given equations.

In fact, after some experimentation, we can see that

$$
\begin{aligned}
f(x)-g(x) & =2(2 f(x)+4 g(x))-3(f(x)+3 g(x)) \\
& =2\left(2 x^{2}+4\right)-3\left(x^{2}+x+6\right) \\
& =x^{2}-3 x-10
\end{aligned}
$$

So to solve $f(x)-g(x)=0$, we solve $x^{2}-3 x-10=0$ or $(x-5)(x+2)=0$. Therefore, $x=5$ or $x=-2$.

Final answer: $5,-2$