[Algebra]
Problem: Let $T=20$. For some real constants $a$ and $b$, the solution sets of the equations $x^{2}+(5 b-T-a) x=T+1$ and $2 x^{2}+(T+8 a-2) x=-10 b$ are the same. Compute $a$.

Solution: Divide each side of the second equation by 2 and equate coefficients to obtain $5 b-T-a=$ $\frac{T}{2}+4 a-1$ and $T+1=-5 b$. Thus $b=\frac{T+1}{-5}$, and plugging this value into the first equation yields $a=-\frac{T}{2}$. With $T=20$, the answer is $\mathbf{- 1 0}$.

Final answer: -10