Which of the options must be true if x satisfies the 2 inequalities gi

Question

Which of the options must be true if x satisfies the two inequalities given below? \frac{5x−1}{x+3}<1 and |2x − 5| ≤ 5 ? A. x < 0 B. x > 0 C. x < 2 D. x > 2 E. None of the above

IanStewart · Accepted Answer

I think it's easiest to start with the second inequality. If |2x - 5| < 5, then, dividing by 2 throughout, |x - 2.5| < 2.5. Since |a - b| is just the distance between a and b on the number line, our inequality says in words "the distance between x and 2.5 is at most 2.5", or in other words, x is between 0 and 5 (inclusive).

Now since we know x is at least 0, we can be sure x+3 is at least 3, and so we can be sure x+3 is positive. So we can now safely multiply both sides of the first inequality by x+3 without worrying about whether we might need to reverse the inequality. Doing that, we find 5x - 1 < x + 3, so 4x < 4, and x < 1. Note here that we assumed in advance that x+3 > 0, or that x > -3, so we haven't even looked at what happens when x < -3 (it turns out in that case there are no solutions, so the complete solution set to the first inequality is -3 < x < 1).

So when we combine the two inequalities, the complete solution set for x is: 0 < x < 1. The question does not, however, ask for the complete solution set for x. Instead it simply asks what must be true. And certainly if x is between 0 and 1, it must be true that x < 2, so C is correct. Answer C would also have been correct if it said "x < 1000" or "-5 < x < 5", or "x is a number"; we're only looking for an answer that must be true when we know 0 < x < 1 is true, and A, B and D can all be false (in the case of A, only because x = 0 is possible), so they're wrong answers, and E is wrong because one answer, C, must indeed be true.

I'd add that if you begin solving by looking at the first inequality rather than at the second, you need to separate it into two cases if you want to multiply by x+3 on both sides, because x+3 might be negative, and if we multiply both sides of an inequality by something negative, we must reverse the inequality. So to analyze the first inequality correctly, we can, as our first case, assume x+3 > 0, or that x > -3, which is what I did above, and from there we find x < 1. Combining the assumption "x > -3" with the solution "x < 1", we learn that -3 < x < 1 is one set of solutions to the inequality. We then must consider the second case, where x + 3 < 0, or where x < -3. You might be able to see by inspection that the inequality won't work in this case (the value of the fraction on the left side will be bigger than 1 for sure), or you can do algebra: we can multiply on both sides by x+3, reversing the inequality when we do because we're multiplying by something negative, to get 5x - 1 > x + 3. Solving, we find x > 1. Since we assumed x < -3, and learned, under that assumption, that x > 1, we've arrived at a contradictory situation, and there can be no solutions in this case, so there are only solutions to the first inequality when x + 3 is positive, and so when -3 < x < 1.

Abhishek009 · Answer

\frac{5x−1}{x+3}<1

Or,  5x - 1 < x + 3

Or,  4x < 4

So,  x < 1

|2x − 5| ≤ 5  can have 2 possibilities

Possibbility - I

2x - 5 ≤ 5

Or,  2x ≤ 10

Or,  x ≤ 5

Possibbility - II

-2x + 5 ≤ 5

Or,  -2x ≤ 0

Or,  0 ≤ 2x

Or,  0 ≤ x

Combining the above we can safely claim that answer must be (C)  x < 2

GMATGuruNY · Answer

Critical points occur when the two sides of an inequality are EQUAL or when the inequality is UNDEFINED.
 \frac{5x−1}{x+3} = 1  when 5x-1 = x+3, with the result that x=1. 
 \frac{5x−1}{x+3}  is undefined when x=-3.
To determine the valid range(s) for x, test one value to the left and one value to the right of each critical point.

x<-3:
Since x=-4 does not satisfy  \frac{5x−1}{x+3} < 1 , this range is not valid.

-3<x<1:
Since x=0 satisfies  \frac{5x−1}{x+3} < 1 , this range is valid.

x>1:
Since x=2 does not satisfy  \frac{5x−1}{x+3} < 1 , this range is not valid.

For the first inequality, the only valid range is -3<x<1.
Since the first inequality requires that x be between -3 and 1, it must be true that x < 2.

C

As worded, the prompt allows us to determine the correct answer without evaluating the second inequality.

Tofupower · Answer

Hi, would you still consider C as the correct answer since x<2 also suggests out of range negative numbers?

Posted from my mobile device

GMATGuruNY · Answer

C does not imply that every value less than 2 will satisfy  \frac{5x-3}{x+1} < 1 .
It implies the reverse:
That every value that satisfies  \frac{5x-3}{x+1} < 1  will be less than 2.

Consider the following DS problems:

Is  \frac{5x-3}{x+1} < 1?

Statement 1: x < 2
In this case, Statement 1 is INSUFFICIENT, since the answer to the question stem is YES if x=0 but NO if x=-10.
In other words:
If x < 2, we cannot conclude that  \frac{5x-3}{x+1} < 1 .

Is x < 2?

Statement 1:  \frac{5x-3}{x+1} < 1 
In this case, Statement 1 is SUFFICIENT, since the answer to the question stem will be YES for every value of x such that  \frac{5x-3}{x+1} < 1 .
In other words:
 If  \frac{5x-3}{x+1} < 1 , we CAN conclude that x < 2.

The statement is green illustrates why C is correct in my earlier post.
Given that  \frac{5x-3}{x+1} < 1 , we can conclude that x < 2.
Put another way:
Given that  \frac{5x-3}{x+1} < 1 , IT MUST BE TRUE that x < 2.

Fdambro294 · Answer

Hard algebra and a “must be true” trap to boot.

The first inequality:

Move everything to the left side and express 1 as = (x + 3) / (x + 3)

We get

(5x - 1)/(x + 3) — (x + 3)/(x + 3) < 0

We have a common denominator so subtract the top, remembering to distribute the minus sign:

(4x - 4) / (x + 3) < 0

The critical point we need to check for our inequality to see which ranges satisfy are:

x = -3 and x = +1

Number line:

[———-(-3)—————(+1)————-]

We have 3 ranges to check to see which ranges satisfy the inequality.

When x > +1 —> inequality above is positive (test x = +2)

And

When x < -3 —> inequality above is positive (test x = -4)

For: -3 < x < 1 ——> the inequality above will be -negative

2nd inequality:

I don’t want to reproduce IanStewart excellent work. It’s where I learned the distance approach in the first place.

Property of absolute values: [aX - a] = [a(x - 1)] = [a] * [x - 1] = +a * [x - 1]

Basically, you can take a common factor from the quantity inside the absolute value bars and then extract it’s positive value.

For the expression:

[2x - 5] = [2 (x - 2.5)] = [2] * [x - 2.5] = 2 * [x - 2.5]

Then if you divide both side of the inequality, you get:

[x - 2.5] </= 2.5

At this point I would read IanStewart discussion of the “distance approach”

We get the following range of X values:

0 </= x </= 5

The first range is more limiting and it is entirely contained within this range, so we have:

-3 < x < 1

Which of the following answers MUST BE TRUE for every possible value that X can take within this range?

x < 0? ——- no, because X can = +(1/2)

x > 0? ——— no, because X can = -1

x > 2? ——— definitely not

Answer C

x < 2

For every value that falls in the range of:

-3 < x < 1

We can say that the X value will always be less than < 2

Answer C

Bunuel

Which of the options must be true if x satisfies the two inequalities given below?

\(\frac{5x−1}{x+3}<1\) and \(|2x − 5| ≤ 5\)?

A. x < 0
B. x > 0
C. x < 2
D. x > 2
E. None of the above

Posted from my mobile device

terminatork07 · Answer

If you think C is correct, just put x = 1 (x < 2). The first inequality doesn't hold that. Therefore answer should be E.

Bunuel · Answer

There is no debate. The answer is C.

(5x - 1)/(x + 3) < 1 and |2x - 5| ≤ 5 give 0 ≤ x < 1 (so x cannot be 1). For any x from this range it would be true to say that x < 2.

P.S. This question belongs to a type of question that tends to be very confusing for many. Therefore, I recommend practicing similar questions from the following collection:  Trickiest Inequality Questions Type: Confusing Ranges .

Kinshook · Answer

Which of the options must be true if x satisfies the two inequalities given below?

\frac{5x−1}{x+3}<1  and  |2x − 5| ≤ 5 ?

\frac{5x−1}{x+3}<1 
 \frac{5x−1}{x+3}-1<0 
 \frac{(5x−1)-(x+3)}{x+3}<0 
 \frac{4x-4}{x+3}<0

Using wavy line method:
-3 < x < 1

|2x − 5| ≤ 5 
-5 ≤ 2x - 5 ≤ 5
0 ≤ 2x ≤ 10
0 ≤ x ≤ 5

Combining we get:
0 ≤ x < 1
or x < 2

IMO C

Which of the options must be true if x satisfies the 2 inequalities gi

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GMAT Problem Solving (PS) Questions

Which of the options must be true if x satisfies the 2 inequalities gi

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