The expression above, where is a constant, can be rewritten in the form for all , where and are constants (that is, neither depends on ). What is the value of ?

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SAT Equivalent Expressions Question 45 | Free SAT Prep | Aniko

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Question 45·Hard·Equivalent Expressions

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\frac{4x-1}{x+2}-\frac{rx+3}{x-1}

The expression above, where $r$ is a constant, can be rewritten in the form

k + \frac{n}{(x+2)(x-1)},

for all $x \ne -2,1$ , where $k$ and $n$ are constants (that is, neither depends on $x$ ). What is the value of $k$ ?

For questions where a rational expression must be rewritten as a constant plus another rational term, first combine everything into a single fraction with a common denominator. Then express the desired form over that same denominator and equate the numerators. Comparing coefficients of $x^2$ , $x$ , and the constant term gives a small system of linear equations in the unknown constants; solve this system carefully, watching for sign errors. This coefficient-comparison method is usually faster and less error-prone than doing full polynomial long division under time pressure.

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Get a common denominator

Rewrite the two fractions as a single fraction. What common denominator can you use for $\dfrac{4x-1}{x+2}$ and $\dfrac{rx+3}{x-1}$ ?

Match the numerators

After you combine the fractions, you will have something of the form $\dfrac{\text{quadratic in } x}{(x+2)(x-1)}$ . Rewrite $k + \dfrac{n}{(x+2)(x-1)}$ as a single fraction with denominator $(x+2)(x-1)$ as well, and then compare the two numerators.

Use coefficient comparison

When two polynomials in $x$ are equal for all $x$ , the coefficients of $x^2$ , $x$ , and the constant term must match. Use this idea to get equations involving $r$ and $k$ , then solve them.

Eliminate r to find k

You will get two equations that both equal $k$ . Set them equal to each other to solve for $r$ , and then plug that value back in to find $k$ .

Desmos Guide

Enter the original expression with a slider for r

In Desmos, type f(x) = (4x - 1)/(x + 2) - (r*x + 3)/(x - 1). Desmos will prompt you to add a slider for r; accept it so you can adjust the value of $r$ .

Enter the target form with sliders for k and n

Type g(x) = k + n/((x + 2)*(x - 1)). Add sliders for k and n when Desmos suggests them. Now you have three sliders: r, k, and n.

Match the two graphs

Adjust the sliders for r, k, and n until the graph of f(x) lies exactly on top of the graph of g(x) for a wide range of $x$ -values (except near $x = -2$ and $x = 1$ , where both have vertical asymptotes). When the graphs coincide everywhere, note the value shown on the k slider—that is the constant $k$ in the rewritten form.

Step-by-step Explanation

Combine the two fractions into one

Start by writing the expression over the common denominator $(x+2)(x-1)$ :

\frac{4x-1}{x+2}-\frac{rx+3}{x-1} = \frac{(4x-1)(x-1) - (rx+3)(x+2)}{(x+2)(x-1)}.

Now expand the numerator:

$(4x-1)(x-1) = 4x^2 - 5x + 1$
$(rx+3)(x+2) = rx^2 + 2rx + 3x + 6$

So the combined numerator is

4x^2 - 5x + 1 - (rx^2 + 2rx + 3x + 6).

Distribute the minus sign:

(4x^2 - 5x + 1) - rx^2 - 2rx - 3x - 6 = (4 - r)x^2 - (8 + 2r)x - 5.

Thus the whole expression is

\frac{(4 - r)x^2 - (8 + 2r)x - 5}{(x+2)(x-1)}.

Match the given target form

We are told the expression can be written as

k + \frac{n}{(x+2)(x-1)}

for all $x \ne -2,1$ .

Write this as a single fraction with denominator $(x+2)(x-1)$ too:

k + \frac{n}{(x+2)(x-1)} = \frac{k(x+2)(x-1) + n}{(x+2)(x-1)}.

Since both forms represent the same expression for all allowed $x$ , their numerators must be equal:

(4 - r)x^2 - (8 + 2r)x - 5 = k(x+2)(x-1) + n.

Expand and compare coefficients

First expand $(x+2)(x-1)$ :

(x+2)(x-1) = x^2 + x - 2.

k(x+2)(x-1) + n = k(x^2 + x - 2) + n = kx^2 + kx - 2k + n.

Now equate the coefficients of matching powers of $x$ from

(4 - r)x^2 - (8 + 2r)x - 5

and

kx^2 + kx - 2k + n.

This gives the system:

$x^2$ terms: $4 - r = k$
$x$ terms: $-(8 + 2r) = k$
constant terms: $-5 = -2k + n$ .

Focus first on the two equations involving $k$ and $r$ :

4 - r = k \quad \text{and} \quad -(8 + 2r) = k.

Solve for r and k

Set the two expressions for $k$ equal to each other:

4 - r = -(8 + 2r).

Solve for $r$ :

4 - r = -8 - 2r \\[0.7em] 4 + r = -8 \\[0.7em] r = -12.

Now substitute $r = -12$ into one of the equations for $k$ , for example $4 - r = k$ :

4 - (-12) = k \\[0.7em] 4 + 12 = k \\[0.7em] 16 = k.

So the required constant is $k = 16$ , which corresponds to answer choice D.

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Step-by-step Explanation

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