If x < 3, is (x + 1)/(x - 3) > 1/3?

Thanks. I have corrected it now. But the steps seem to be fine.

Given: x < 3

Target question: Is (x + 1)/(x - 3) > 1/3?
This is a good candidate for rephrasing the target question.
Aside: See below for a video with tips on rephrasing the target question

Since x < 3, we know that (x - 3) will always be NEGATIVE.
So, let's take (x + 1)/(x - 3) > 1/3, and multiply both sides by (x - 3)
We get: x + 1 < (1/3)(x - 3) [ASIDE: since we multiplied both sides by a NEGATIVE value, we REVERSED the inequality sign]
Now take x + 1 < (1/3)(x - 3) and multiply both sides by 3 to get: 3(x + 1) < x - 3
Expand to get: 3x + 3 < x - 3
Subtract x from both sides: 2x + 3 < -3
Subtract 3 from both sides: 2x < -6
Divide both sides by 2 to get: x < -3

This equivalent inequality is much easier to work with. So, ....
REPHRASED target question: Is x < -3?

Statement 1: x < 2
There are several values of x that satisfy statement 1. Here are two:
Case a: x = -7, in which case x < -3
Case b: x = 0, in which case x > -3
Since we cannot answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: x > -1
If x is greater than -1, then we can be 100% certain that x IS NOT less than -3
Since we can answer the REPHRASED target question with certainty, statement 2 is SUFFICIENT

Answer: B

RELATED VIDEOS FROM OUR COURSE

Step 1-
(x-3)+4
X-3
gives
1+4/(x-3)>1/3
(2/(x-3))>-1/3
Now, option b , since x>-1 & <3,LHS can never be greater than -1/3
So B is sufficient

\(\frac{x+1}{x-3}>1/3…\frac{(x+1)3-(x-3)1}{3(x-3)}>0…\frac{3x+3-x+3}{3x-9}>0…\frac{2(x+3)}{3(x-3)}>0\)
\(\frac{2(x+3)}{3(x-3)}>0…\frac{(x+3)}{x-3}>0…true:x<-3\)
\((ie).x=-7:\frac{2(-7+3)}{3(-7-3)}>0…\frac{2(-4)}{3(-10)}>0…\frac{-8}{-30}>0\)

(1) x < 2 insufic

(2) x > -1 sufic

Ans (B)

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