2030 AMC 8 Problems/Problem 1

Problem

The equation is shown as: $\frac{24x^2+25x-47}{ax-2}=-8x-3-\frac{53}{ax-2}$ is true for all values except when $x=\frac{2}{a}$ , where a is constant. What is the value of a?

$\text {(A)}\ -16 \qquad \text {(B)}\ -3 \qquad \text {(C)}\ 0 \qquad \text {(D)}\ 3 \qquad \text {(E)} 16$

Solution

The faster way is to multiply each side of the given equation by 𝑎𝑥−2 (so you can get rid of the fraction). When you multiply each side by 𝑎𝑥−2, you should have: $24x^2+25x-47=(-8x-3)(ax-2)-53$ Using the FOIL method, you should then multiply (−8𝑥−3) and (𝑎𝑥−2). After that you should have the following: $24x^2+25x-47=-8ax^2-3ax+16x+6-53$ Then, reduce on the right side of the equation: $24x^2+25x-47=-8ax^2-3ax+16x-47$ Since the coefficients of the x² term has to be equal on both sides of the equation, −8a=24, and that concludes us with 𝑎=−3. $\Rightarrow\boxed{\mathrm{(B)}\ -3}$ .

2030 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 2
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All AJHSME/AMC 8 Problems and Solutions

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2030 AMC 8 Problems/Problem 1

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