Question 135·Hard·Advanced Math

The two forms agree for every allowed $x$, so write their equality: $\frac{4x-1}{x+2}-\frac{rx+3}{x-1}=k+\frac{n}{(x+2)(x-1)}$. This is an identity, not an equation to solve for one value of $x$.

Since $x\ne-2,1$, you can multiply by the whole common denominator $(x+2)(x-1)$: $(4x-1)(x-1)-(rx+3)(x+2)=k(x+2)(x-1)+n$. The $k$ term becomes $k$ times the denominator; it does not disappear.

Each side of the cleared equation is at most a quadratic, a polynomial with no power higher than $x^2$. A nonzero quadratic cannot equal zero at three distinct inputs. So matching at three allowed inputs is enough to establish this identity. Use $2$, $3$, and $4$; none makes a denominator zero.

Type $x_1=[2,3,4]$, then type the original equality with each $x$ changed to $x_1$ and $=$ changed to $\sim$. The subscript makes $x_1$ a list of inputs rather than a graph variable; $\sim$ tells Desmos to fit $r$, $k$, and $n$. Under PARAMETERS, Desmos shows $r=-12$, $k=16$, and $n=27$. So the constant in the requested form is $16$. Choice D.

Write the first numerator as a multiple of its denominator plus a remainder: $4x-1=4(x+2)-9$. Divide both parts by $x+2$: $\frac{4x-1}{x+2}=4-\frac{9}{x+2}$.

Do the same with the second denominator: $rx+3=r(x-1)+(r+3)$. Divide both parts by $x-1$: $\frac{rx+3}{x-1}=r+\frac{r+3}{x-1}$.

The original expression subtracts the entire second fraction, so substituting the two split forms gives: $4-r-\frac{9}{x+2}-\frac{r+3}{x-1}$. The constant part is $4-r$; the two remaining fractions still need to become one.

Give the leftover fractions the requested denominator $(x+2)(x-1)$: $-\frac{9}{x+2}-\frac{r+3}{x-1}=\frac{-9(x-1)-(r+3)(x+2)}{(x+2)(x-1)}$. Each numerator gets the factor its denominator was missing.

Distribute in the new numerator, keeping the minus sign before the second product: $\begin{aligned}-9(x-1)-(r+3)(x+2)&=-9x+9-rx-2r-3x-6\\&=-(r+12)x+3-2r.\end{aligned}$

The requested numerator $n$ is a constant, so its coefficient of $x$ must be $0$: $-(r+12)=0$. Solve for $r$: $r=-12$.

The constant part was $4-r$, so substitute $r=-12$: $k=4-(-12)=16$. The value of $k$ is $16$. Choice D.