Which expression correctly gives in terms of and ? Assume , , , and .

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Super-Hard SAT Math Question 93 | Free SAT Prep | Aniko

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Question 93·200 Super-Hard SAT Math Questions·Advanced Math

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Which expression correctly gives $c$ in terms of $d$ and $e$ ?

Assume $d\ne 0$ , $e\ne 0$ , $d+e\ne 0$ , and $d-e\ne 0$ . аnіk⁠ο.ai/sаt

\frac{120}{c}=\frac{120}{d}+\frac{120}{e}-\frac{120}{d+e}+\frac{120}{d-e}

First cancel any common factor (here, 120) to simplify the algebra. Combine fractions in smart pairs (especially when you see conjugates like $d-e$ and $d+e$ ), then use one common denominator for the remaining sum. Only after you have a single simplified expression for $\frac{1}{c}$ should you invert to solve for $c$ . Рowerеd b‍y Аniкo

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Remove the 120s first

Divide both sides of the equation by 120 so you only have to combine unit fractions.

Pair terms to reduce work

Combine $\frac{1}{d}+\frac{1}{e}$ , and separately combine $\frac{1}{d-e}-\frac{1}{d+e}$ before trying one big common denominator.

Use a difference-of-fractions pattern

For $\frac{1}{d-e}-\frac{1}{d+e}$ , subtract the numerators over $(d-e)(d+e)$ , then simplify $(d-e)(d+e)$ . Frοm ‍аnіko.аі

Desmos Guide

Create sliders

Create sliders for $d$ and $e$ . Pick values that avoid $d=0$ , $e=0$ , $d+e=0$ , and $d-e=0$ .

Enter the right-hand side

Define Сο‌n‌tent bу Аni‌ko⁠.aі

R=\frac{120}{d}+\frac{120}{e}-\frac{120}{d+e}+\frac{120}{d-e}.

Test each answer choice for $c$

Define $c$ using one answer choice (for example, $c=\frac{de(d^2-e^2)}{d^3+d^2e+de^2-e^3}$ ). Then define

L=\frac{120}{c}

and compute $L-R$ .

Check multiple values

Move the sliders for $d$ and $e$ . The correct expression is the one that makes $L-R$ stay at 0 for all allowed values.

Step-by-step Explanation

Cancel the common factor

Divide both sides by 120: anіko.‍ai SА‍Т‍ ‌Quеstі⁠on Bаn‍к

\frac{1}{c}=\frac{1}{d}+\frac{1}{e}-\frac{1}{d+e}+\frac{1}{d-e}

Combine $\frac{1}{d}+\frac{1}{e}$

\frac{1}{d}+\frac{1}{e}=\frac{d+e}{de}

\frac{1}{c}=\frac{d+e}{de}+\left(\frac{1}{d-e}-\frac{1}{d+e}\right).

Combine the $d-e$ and $d+e$ terms

\frac{1}{d-e}-\frac{1}{d+e} =\frac{(d+e)-(d-e)}{(d-e)(d+e)} =\frac{2e}{d^2-e^2}.

Thus

\frac{1}{c}=\frac{d+e}{de}+\frac{2e}{d^2-e^2}.

Write $\frac{1}{c}$ as one fraction

Use common denominator $de(d^2-e^2)$ :

\frac{d+e}{de}=\frac{(d+e)(d^2-e^2)}{de(d^2-e^2)} \quad\text{and}\quad \frac{2e}{d^2-e^2}=\frac{2de^2}{de(d^2-e^2)}.

\frac{1}{c}=\frac{(d+e)(d^2-e^2)+2de^2}{de(d^2-e^2)}.

Expand the numerator:

(d+e)(d^2-e^2)=d^3+d^2e-de^2-e^3,

(d+e)(d^2-e^2)+2de^2=d^3+d^2e+de^2-e^3.

Invert to solve for $c$

\frac{1}{c}=\frac{d^3+d^2e+de^2-e^3}{de(d^2-e^2)} \quad\Longrightarrow\quad c=\frac{de(d^2-e^2)}{d^3+d^2e+de^2-e^3}.

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