[Algebra]
Problem: Suppose that $x$ and $y$ are real numbers with $3 x+4 y=10$. Determine the minimum possible value of $x^{2}+16 y^{2}$.

Solution: Since $3 x+4 y=10$, then $4 y=10-3 x$.

Therefore, when $3 x+4 y=10$,

$$
\begin{aligned}
x^{2}+16 y^{2} & =x^{2}+(4 y)^{2} \\
& =x^{2}+(10-3 x)^{2} \\
& =x^{2}+\left(9 x^{2}-60 x+100\right) \\
& =10 x^{2}-60 x+100 \\
& =10\left(x^{2}-6 x+10\right) \\
& =10\left(x^{2}-6 x+9+1\right) \\
& =10\left((x-3)^{2}+1\right) \\
& =10(x-3)^{2}+10
\end{aligned}
$$

Since $(x-3)^{2} \geq 0$, then the minimum possible value of $10(x-3)^{2}+10$ is $10(0)+10=10$. This occurs when $(x-3)^{2}=0$ or $x=3$.

Therefore, the minimum possible value of $x^{2}+16 y^{2}$ when $3 x+4 y=10$ is 10 .

Final answer: 10