[Algebra]
Problem: Let $T=\frac{1}{40}$. Given that $x, y$, and $z$ are real numbers such that $x+y=5, x^{2}-y^{2}=\frac{1}{T}$, and $x-z=-7$, compute $x+z$

Solution: Note that $x^{2}-y^{2}=(x+y)(x-y)=5(x-y)$, hence $x-y=\frac{1}{5 T}$. Then $x+z=(x+y)+(x-y)+(z-x)=$ $5+\frac{1}{5 T}+7=12+\frac{1}{5 T}$. With $T=\frac{1}{40}$, the answer is thus $12+8=\mathbf{2 0}$.

Final answer: 20