Question 215·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

Choose $t=1$ as a sign check; it doesn't make the given denominator zero. Type $t=1$, then $u=\frac{t-5}{2t+1}$ in Desmos. It shows $u=-\frac43$ (tap the fraction button). Your final formula should return $t=1$ for this value of $u$.

The problem says $2t+1\ne0$, so multiply both whole sides by $2t+1$ to clear the fraction: $u(2t+1)=t-5$.

Distribute $u$ across both terms in the parentheses: $2ut+u=t-5$. Now $t$ appears twice, so you can't isolate it by dividing yet.

Subtract $2ut$ from both sides: $u=t-2ut-5$. The two terms containing $t$ are now on the right.

Add $5$ to both sides: $u+5=t-2ut$. Keep the plus sign on $5$; moving a term across the equals sign changes its sign.

Use factoring, the reverse of distribution, to pull $t$ out of both terms on the right: $u+5=t(1-2u)$. When the target appears twice, collect its terms and factor it out before dividing.

Divide by $1-2u$: $t=\frac{5+u}{1-2u}$. This divisor can't be zero: $u=\frac12$ would make the right side of the previous line zero but leave $u+5$ nonzero. Type $\frac{5+u}{1-2u}$ in Desmos; it shows $1$ for the $u=-\frac43$ found earlier, matching the $t=1$ you chose. So this equation expresses $t$ in terms of $u$. Choice D.