Question

The equation

$\frac{24x^{2} +25x-47}{ax-2} = -8x - 3-\frac{53}{ax-2}$

is true for all values of x≠ $\frac{2}{a}$

, where a is a constant.

What is the value of a?

A) -16
B) -3
C) 3
D) 16

Answer 1

Answer:

Step-by-step explanation:

There are two ways to solve this question. 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:

24x²+25x−47=(−8x−3)(ax−2)−53

You should then multiply (−8x−3) and (ax−2) using FOIL.

24x²+25x−47=−8ax²−3ax+16x+6−53

Then, reduce on the right side of the equation

24x²+25x−47=−8ax²−3ax+16x−47

Since the coefficients of the x²-term have to be equal on both sides of the equation, −8a=24, or a=−3.

The other option which is longer and more tedious is to attempt to plug in all of the answer choices for a and see which answer choice makes both sides of the equation equal. Again, this is the longer option, and I do not recommend it for the actual SAT as it will waste too much time.

The final answer is B.

Answer 2

ᴛʜᴇ ᴀɴꜱᴡᴇʀ ɪꜱ ɪɴ ᴛʜᴇ ᴀᴛᴛᴀᴄʜᴇᴅ ꜰɪʟᴇ

ᴛʜᴇ ᴄᴏʀʀᴇᴄᴛ ᴀɴꜱᴡᴇʀ ɪꜱ ʙ

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