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Tricky simplification

Expert replies
by Brent@GMATPrepNow » Mon Dec 05, 2016 8:40 am
If a + c ≠ b, then [4a²−(a + b − c)²]/[a − b + c] =

A) a - b - c
B) a + b - c
C) 3a - b - c
D) 3a + b + c
E) 3a + b - c

Answer: E
Brent Hanneson - Creator of GMATPrepNow.com
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Source: — Problem Solving |

by regor60 » Mon Dec 05, 2016 10:13 am
Brent@GMATPrepNow wrote:If a + c ≠ b, then [4a²−(a + b − c)²]/[a − b + c] =

A) a - b - c
B) a + b - c
C) 3a - b - c
D) 3a + b + c
E) 3a + b - c

Answer: E
Pick some easy numbers:

a=1,b=2,c=3

Plug into original equation with the result = 2

Sub same values into the answer choices and see which yields 2:

A. Result -4 fails

B. " 0 fails

C. " -2 fails

D. " 8 fails

E. " 2 Winner
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by regor60 » Mon Dec 05, 2016 10:13 am
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by Brent@GMATPrepNow » Tue Dec 06, 2016 10:36 am
Brent@GMATPrepNow wrote:If a + c ≠ b, then [4a²−(a + b − c)²]/[a − b + c] =

A) a - b - c
B) a + b - c
C) 3a - b - c
D) 3a + b + c
E) 3a + b - c

Answer: E
Here's an algebraic approach
First recognize that 4a² = (2a)²
So, the numerator can be written as (2a)² − (a + b − c)²
This happens to be a difference of squares
So, we can factor the the numerator as follows: [2a + (a + b − c)][2a - (a + b − c)]
Simplify to get: [3a + b − c][a - b + c)]

So, the given fraction becomes [3a + b − c][a - b + c)]/[a − b + c]
This simplifies to be: 3a + b − c
Answer: E

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Matt@VeritasPrep » Thu Dec 08, 2016 8:03 pm
Another approach!

Let's say that

[4a²-(a + b - c)²]/[a - b + c] = x

4a² - (a + b - c)² = (a - b + c) * x

We know that 4a² - a² on the left is going to give us 3a², so we must have a 3a term on the right hand side.

We know that - (-c)² is going to give us -c², so we must have +c * -c, or -c on the right hand side.

We know that - (+b)² is going to give us -b², so we must have -b * +b, or +b on the right hand side.

From there, our answer can only be 3a + b - c!
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by Matt@VeritasPrep » Thu Dec 08, 2016 8:04 pm
My answer above is obviously much less cool (and less useful!) than Brent's, but I did want to show a way to bash your way out of the problem under test conditions if you're stuck. (Alas, the elegant answer doesn't always arrive in time!)
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