SAT Question 433184f1: Equivalent Expressions answer and explanation with Desmos solution

Choice D is correct. The expression $\frac{4}{4x-5}-\frac{1}{x+1}$ can be rewritten as $\frac{4}{4x-5}+\frac{\left(-1\right)}{x+1}$. To add the two terms of this expression, the terms can be rewritten with a common denominator. Since $\frac{x+1}{x+1}=1$, the expression $\frac{4}{4x-5}$ can be rewritten as $\frac{\left(x+1\right)\left(4\right)}{\left(x+1\right)\left(4x-5\right)}$. Since $\frac{4x-5}{4x-5}=1$, the expression $\frac{-1}{x+1}$ can be rewritten as $\frac{\left(4x-5\right)\left(-1\right)}{\left(4x-5\right)\left(x+1\right)}$. Therefore, the expression $\frac{4}{4x-5}+\frac{\left(-1\right)}{x+1}$ can be rewritten as $\frac{\left(x+1\right)\left(4\right)}{\left(x+1\right)\left(4x-5\right)}+\frac{\left(4x-5\right)\left(-1\right)}{\left(4x-5\right)\left(x+1\right)}$, which is equivalent to $\frac{\left(x+1\right)\left(4\right)+\left(4x-5\right)\left(-1\right)}{\left(x+1\right)\left(4x-5\right)}$. Applying the distributive property to each term of the numerator yields $\frac{\left(4x+4\right)+\left(-4x+5\right)}{\left(x+1\right)\left(4x-5\right)}$, or $\frac{\left(4x+\left(-4x\right)\right)+\left(4+5\right)}{\left(x+1\right)\left(4x-5\right)}$. Adding like terms in the numerator yields $\frac{9}{\left(x+1\right)\left(4x-5\right)}$.