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ramzin
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What range of values of x will satisfy the inequality |2x + Updated on: 28 Apr 2014, 02:06
Originally posted by ramzin on 25 May 2011, 07:04.
Last edited by Bunuel on 28 Apr 2014, 02:06, edited 2 times in total.
Renamed the topic, edited the question and the OA.
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C
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71% (02:31) correct 29% (02:37) wrong based on 1897 sessions

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What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

A. x < -1/9 or x > 5

B. -1 < x < 1/9

C. -1/9 < x < 1

D. -1/9 < x < 5

E. x < -1/9 or x > 1
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C
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What range of values of x will satisfy the inequality |2x + 28 Apr 2014, 02:06
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ramzin
What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

A. x < -1/9 or x > 5

B. -1 < x < 1/9

C. -1/9 < x < 1

D. -1/9 < x < 5

E. x < -1/9 or x > 1
Show more

Since both sides of the inequality are non-negative we can safely square:

\(4x^2+12x+9>49x^2-28x+4\);

\(45x^2-40x-5<0\);

\(9x^2-8x-1<0\);

\((x+\frac{1}{9})(x-1)<0\).

The "roots" are -1/9 and 1 (Solving Quadratic Inequalities: https://gmatclub.com/forum/solving-quad ... 70528.html). "<" sign indicates that the solution must be between the roots: \(-\frac{1}{9}<x<1\).

Answer: C.
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What range of values of x will satisfy the inequality |2x + 25 May 2011, 07:26
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two methods here.

1. using the solutions here. x= -3/2 and x= 2/7 are solutions.
check for the regions

a. x<-3/2, b. -3/2<x<2/7 and c. x>2/7

values are a -(2x+3) > (7x-2) giving x <-1/9 hence not a solution.

b (2x+3) > (7x-2) giving x < 1 a solution.

c (2x+3) > -(7x-2) giving x>-1/9 hence the solution is -1/9 < x < 1

or

2. squaring both sides

(2x+3) ^2 > (7x-2)^2

gives 9x^2 -8x - 1 > 0 giving solution -1/9 < x < 1.

Hence C you can use whichever you are comfortable with.
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gauravsoni
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What range of values of x will satisfy the inequality |2x + 28 Apr 2014, 04:27
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Bunuel
ramzin
What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

A. x < -1/9 or x > 5

B. -1 < x < 1/9

C. -1/9 < x < 1

D. -1/9 < x < 5

E. x < -1/9 or x > 1

Since both sides of the inequality are non-negative we can safely square:

\(4x^2+12x+9>49x^2-28x+4\) --> \(45x^2-40x-5<0\) --> \(9x^2-8x-1<0\) --> \((x+\frac{1}{9})(x-1)<0\).

The "roots" are -1/9 and 1 (Solving Quadratic Inequalities: solving-quadratic-inequalities-graphic-approach-170528.html). "<" sign indicates that the solution must be between the roots: \(-\frac{1}{9}<x<1\).

Answer: C.
Show more

Hi Bunuel ,

Thanks for the answer. I did not understand how are the roots -1/9 and 1. I am getting the roots as 9 and 1. Can you please elaborate.
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What range of values of x will satisfy the inequality |2x + 28 Apr 2014, 04:38
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gauravsoni
Bunuel
ramzin
What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

A. x < -1/9 or x > 5

B. -1 < x < 1/9

C. -1/9 < x < 1

D. -1/9 < x < 5

E. x < -1/9 or x > 1

Since both sides of the inequality are non-negative we can safely square:

\(4x^2+12x+9>49x^2-28x+4\) --> \(45x^2-40x-5<0\) --> \(9x^2-8x-1<0\) --> \((x+\frac{1}{9})(x-1)<0\).

The "roots" are -1/9 and 1 (Solving Quadratic Inequalities: solving-quadratic-inequalities-graphic-approach-170528.html). "<" sign indicates that the solution must be between the roots: \(-\frac{1}{9}<x<1\).

Answer: C.

Hi Bunuel ,

Thanks for the answer. I did not understand how are the roots -1/9 and 1. I am getting the roots as 9 and 1. Can you please elaborate.
Show more

\((x+\frac{1}{9})(x-1)=0\) --> \(x+\frac{1}{9}=0\) or \(x-1=0\) --> \(x=-\frac{1}{9}\) or \(x=1\).
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What range of values of x will satisfy the inequality |2x + 28 Apr 2014, 05:45
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Bunuel
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ramzin
What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

A. x < -1/9 or x > 5

B. -1 < x < 1/9

C. -1/9 < x < 1

D. -1/9 < x < 5

E. x < -1/9 or x > 1

Since both sides of the inequality are non-negative we can safely square:

\(4x^2+12x+9>49x^2-28x+4\) --> \(45x^2-40x-5<0\) --> \(9x^2-8x-1<0\) --> \((x+\frac{1}{9})(x-1)<0\).

The "roots" are -1/9 and 1 (Solving Quadratic Inequalities: solving-quadratic-inequalities-graphic-approach-170528.html). "<" sign indicates that the solution must be between the roots: \(-\frac{1}{9}<x<1\).

Answer: C.

Hi Bunuel ,

Thanks for the answer. I did not understand how are the roots -1/9 and 1. I am getting the roots as 9 and 1. Can you please elaborate.
Show more

\((x+\frac{1}{9})(x-1)=0\) --> \(x+\frac{1}{9}=0\) or \(x-1=0\) --> \(x=-\frac{1}{9}\) or \(x=1\).[/quote]

That part is fine , i'm solving the quadratic equation 9x^2-8x-1 < 0 as (x-9)(x+1) then getting x = 9 , x = -1. I also check out your link for solving quadratic equations in equalities but could relate it.
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What range of values of x will satisfy the inequality |2x + 28 Apr 2014, 05:52
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Bunuel
Bunuel
What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

A. x < -1/9 or x > 5

B. -1 < x < 1/9

C. -1/9 < x < 1

D. -1/9 < x < 5

E. x < -1/9 or x > 1

Since both sides of the inequality are non-negative we can safely square:

\(4x^2+12x+9>49x^2-28x+4\) --> \(45x^2-40x-5<0\) --> \(9x^2-8x-1<0\) --> \((x+\frac{1}{9})(x-1)<0\).

The "roots" are -1/9 and 1 (Solving Quadratic Inequalities: solving-quadratic-inequalities-graphic-approach-170528.html). "<" sign indicates that the solution must be between the roots: \(-\frac{1}{9}<x<1\).

Answer: C.

Hi Bunuel ,

Thanks for the answer. I did not understand how are the roots -1/9 and 1. I am getting the roots as 9 and 1. Can you please elaborate.

\((x+\frac{1}{9})(x-1)=0\) --> \(x+\frac{1}{9}=0\) or \(x-1=0\) --> \(x=-\frac{1}{9}\) or \(x=1\).
Show more

That part is fine , i'm solving the quadratic equation 9x^2-8x-1 < 0 as (x-9)(x+1) then getting x = 9 , x = -1. I also check out your link for solving quadratic equations in equalities but could relate it.[/quote]

(x-9)(x+1) is not a correct factoring of 9x^2-8x-1, it should be \((x+\frac{1}{9})(x-1)\) (\((9x+1)(x-1)\)).

Factoring Quadratics: https://www.purplemath.com/modules/factquad.htm

Solving Quadratic Equations: https://www.purplemath.com/modules/solvquad.htm

Hope this helps.
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What range of values of x will satisfy the inequality |2x + 29 Apr 2014, 01:49
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amit2k9
two methods here.

1. using the solutions here. x= -3/2 and x= 2/7 are solutions.
check for the regions

a. x<-3/2, b. -3/2<x<2/7 and c. x>2/7

values are a -(2x+3) > (7x-2) giving x <-1/9 hence not a solution.

b (2x+3) > (7x-2) giving x < 1 a solution.

c (2x+3) > -(7x-2) giving x>-1/9 hence the solution is -1/9 < x < 1

or

2. squaring both sides

(2x+3) ^2 > (7x-2)^2

gives 9x^2 -8x - 1 > 0 giving solution -1/9 < x < 1.

Hence C you can use whichever you are comfortable with.
Show more

Hi Bunuel,
Thanks for the solution! I get the squaring both sides approach, but it takes over 3 minutes for me to do it that way.
Is there a faster way to solve this?

I noticed Amit provided another method, but I'm not sure I understand the first approach completely. Could you explain why he used the "-" sign in the first and third parts (highlighted in blue above)?
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What range of values of x will satisfy the inequality |2x + 29 Apr 2014, 04:46
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gauravsoni
Bunuel
ramzin
What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

A. x < -1/9 or x > 5

B. -1 < x < 1/9

C. -1/9 < x < 1

D. -1/9 < x < 5

E. x < -1/9 or x > 1

Since both sides of the inequality are non-negative we can safely square:

\(4x^2+12x+9>49x^2-28x+4\) --> \(45x^2-40x-5<0\) --> \(9x^2-8x-1<0\) --> \((x+\frac{1}{9})(x-1)<0\).

The "roots" are -1/9 and 1 (Solving Quadratic Inequalities: solving-quadratic-inequalities-graphic-approach-170528.html). "<" sign indicates that the solution must be between the roots: \(-\frac{1}{9}<x<1\).

Answer: C.

Hi Bunuel ,

Thanks for the answer. I did not understand how are the roots -1/9 and 1. I am getting the roots as 9 and 1. Can you please elaborate.
Show more

Hi Gauravsoni,

\(9x^2-8x-1<0\)

Can be factored 9x^2 -9x+x-1 <0

9x(x-1)+1 (x-1) <0 or (9x+1)(x-1)<0 or x=-1/9 or x=1
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What range of values of x will satisfy the inequality |2x + 07 May 2014, 10:41
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hi bunnel

what do u mean when u say the following:

Quote:
Since both sides of the inequality are non-negative we can safely square:

4x^2+12x+9>49x^2-28x+4 --> 45x^2-40x-5<0 --> 9x^2-8x-1<0 --> (x+\frac{1}{9})(x-1)<0.
Show more

how how we solve inequality which is negative??
whenever we see absolute values on each side of the sign... should we use the squaring approach??
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What range of values of x will satisfy the inequality |2x + 08 May 2014, 03:24
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nandinigaur
hi bunnel

what do u mean when u say the following:

Quote:
Since both sides of the inequality are non-negative we can safely square:

4x^2+12x+9>49x^2-28x+4 --> 45x^2-40x-5<0 --> 9x^2-8x-1<0 --> (x+\frac{1}{9})(x-1)<0.

how how we solve inequality which is negative??
whenever we see absolute values on each side of the sign... should we use the squaring approach??
Show more

We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality)

Adding/subtracting/multiplying/dividing inequalities: help-with-add-subtract-mult-divid-multiple-inequalities-155290.html

As for your other questions: it depends on a question.
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What range of values of x will satisfy the inequality |2x + 12 May 2014, 00:13
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Bunuel
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What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

A. x < -1/9 or x > 5

B. -1 < x < 1/9

C. -1/9 < x < 1

D. -1/9 < x < 5

E. x < -1/9 or x > 1

Since both sides of the inequality are non-negative we can safely square:

\(4x^2+12x+9>49x^2-28x+4\) --> \(45x^2-40x-5<0\) --> \(9x^2-8x-1<0\) --> \((x+\frac{1}{9})(x-1)<0\).

The "roots" are -1/9 and 1 (Solving Quadratic Inequalities: solving-quadratic-inequalities-graphic-approach-170528.html). "<" sign indicates that the solution must be between the roots: \(-\frac{1}{9}<x<1\).

Answer: C.
Show more


If we put in the value x = 0 , the inequality is not being satisfied , even though it is within the range. Am I missing something here ?..

I tried to deduce the answers by substituting possible values from the various option ranges.
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What range of values of x will satisfy the inequality |2x + 12 May 2014, 00:18
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Bunuel
ramzin
What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

A. x < -1/9 or x > 5

B. -1 < x < 1/9

C. -1/9 < x < 1

D. -1/9 < x < 5

E. x < -1/9 or x > 1

Since both sides of the inequality are non-negative we can safely square:

\(4x^2+12x+9>49x^2-28x+4\) --> \(45x^2-40x-5<0\) --> \(9x^2-8x-1<0\) --> \((x+\frac{1}{9})(x-1)<0\).

The "roots" are -1/9 and 1 (Solving Quadratic Inequalities: solving-quadratic-inequalities-graphic-approach-170528.html). "<" sign indicates that the solution must be between the roots: \(-\frac{1}{9}<x<1\).

Answer: C.


If we put in the value x = 0 , the inequality is not being satisfied , even though it is within the range. Am I missing something here ?..

I tried to deduce the answers by substituting possible values from the various option ranges.
Show more

If x = 0, then |2x + 3| = 3 and |7x - 2| = 2 --> 3 > 2. So, 0 is a possible value of x.

Hope it helps.
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What range of values of x will satisfy the inequality |2x + 28 Mar 2016, 08:03
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What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

Plotting the The halfs of the equation we get the diagram as shown...shaded portion shows the area of interest satisfyin the above inequality.
In upper half x=1 is the point of intersection i.e the maximum value of x. Only C satisfies.

A. x < -1/9 or x > 5

B. -1 < x < 1/9

C. -1/9 < x < 1

D. -1/9 < x < 5

E. x < -1/9 or x > 1
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What range of values of x will satisfy the inequality |2x + 12 Nov 2023, 15:44
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What is wrong with the following in the negative scenario B below?

|2x+3| > |7x-2|

(A)
2x + 3 > 7x -2
2x - 7x > -5
-5x > -5
x < -5 / -5
x < 1

(B)
-(2x+3) > 7x-2
2x + 3 < -1 (7x-2)
2x + 3 < -7x + 2
2x + 7x < 2 - 3
9x < -1
x < -1/9 ?

This doesn't match ans choice C
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What range of values of x will satisfy the inequality |2x + 13 Nov 2023, 00:22
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What is wrong with the following in the negative scenario B below?

|2x+3| > |7x-2|

(A)
2x + 3 > 7x -2
2x - 7x > -5
-5x > -5
x < -5 / -5
x < 1

(B)
-(2x+3) > 7x-2
2x + 3 < -1 (7x-2)
2x + 3 < -7x + 2
2x + 7x < 2 - 3
9x < -1
x < -1/9 ?

This doesn't match ans choice C
Show more

What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

A. x < -1/9 or x > 5

B. -1 < x < 1/9

C. -1/9 < x < 1

D. -1/9 < x < 5

E. x < -1/9 or x > 1

The critical points (also known as transition points or key points) for |2x + 3| and |7x - 2| are x = -3/2 and x = 2/7, respectively. These critical points are where the expressions inside the modulus equal zero, signifying a transition from negative to positive values (or vice-versa).

Considering the two critical points, we analyze three ranges:

1. If x < -3/2, then 2x + 3 < 0 and 7x - 2 < 0. Thus, |2x + 3| = -(2x + 3) and |7x - 2| = -(7x - 2). Hence, for this range we'd get: -(2x + 3) > -(7x - 2), which gives x > 1. Discard this range because it contradicts with the range we consider: x < -3/2.

2. If -3/2 ≤ x ≤ 2/7, then 2x + 3 ≥ 0 and 7x - 2 ≤ 0. Thus, |2x + 3| = 2x + 3 and |7x - 2| = -(7x - 2). Hence, for this range we'd get: 2x + 3 > -(7x - 2), which gives x > -1/9. Combining with the range we consider, we'd get -1/9 < x ≤ 2/7.

3. If x > 2/7, then 2x + 3 > 0 and 7x - 2 > 0. Thus, |2x + 3| = 2x + 3 and |7x - 2| = 7x - 2. Hence, for this range we'd get: 2x + 3 > 7x - 2, which gives x < 1. Combining with the range we consider, we'd get 2/7 ≤ x < 1.

So, we get two ranges: -1/9 < x ≤ 2/7 and 2/7 ≤ x < 1, which give: -1/9 < x < 1.

Answer: C.

10. Absolute Value



For more check Ultimate GMAT Quantitative Megathread



Hope it helps.
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What range of values of x will satisfy the inequality |2x + 02 Jul 2024, 11:35
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Here is my explanation to it. Hope you like it.

ramzin
What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

A. x < -1/9 or x > 5

B. -1 < x < 1/9

C. -1/9 < x < 1

D. -1/9 < x < 5

E. x < -1/9 or x > 1
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Posted from my mobile device
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lnyngayan
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What range of values of x will satisfy the inequality |2x + 02 Sep 2024, 06:44
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Bunuel
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What is wrong with the following in the negative scenario B below?

|2x+3| > |7x-2|

(A)
2x + 3 > 7x -2
2x - 7x > -5
-5x > -5
x < -5 / -5
x < 1

(B)
-(2x+3) > 7x-2
2x + 3 < -1 (7x-2)
2x + 3 < -7x + 2
2x + 7x < 2 - 3
9x < -1
x < -1/9 ?

This doesn't match ans choice C

What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

A. x < -1/9 or x > 5

B. -1 < x < 1/9

C. -1/9 < x < 1

D. -1/9 < x < 5

E. x < -1/9 or x > 1

The critical points (also known as transition points or key points) for |2x + 3| and |7x - 2| are x = -3/2 and x = 2/7, respectively. These critical points are where the expressions inside the modulus equal zero, signifying a transition from negative to positive values (or vice-versa).

Considering the two critical points, we analyze three ranges:

1. If x < -3/2, then 2x + 3 < 0 and 7x - 2 < 0. Thus, |2x + 3| = -(2x + 3) and |7x - 2| = -(7x - 2). Hence, for this range we'd get: -(2x + 3) > -(7x - 2), which gives x > 1. Discard this range because it contradicts with the range we consider: x < -3/2.

2. If -3/2 ≤ x ≤ 2/7, then 2x + 3 ≥ 0 and 7x - 2 ≤ 0. Thus, |2x + 3| = 2x + 3 and |7x - 2| = -(7x - 2). Hence, for this range we'd get: 2x + 3 > -(7x - 2), which gives x > -1/9. Combining with the range we consider, we'd get -1/9 < x ≤ 2/7.

3. If x > 2/7, then 2x + 3 > 0 and 7x - 2 > 0. Thus, |2x + 3| = 2x + 3 and |7x - 2| = 7x - 2. Hence, for this range we'd get: 2x + 3 > 7x - 2, which gives x < 1. Combining with the range we consider, we'd get 2/7 ≤ x < 1.

So, we get two ranges: -1/9 < x ≤ 2/7 and 2/7 ≤ x < 1, which give: -1/9 < x < 1.

Answer: C.

10. Absolute Value



For more check Ultimate GMAT Quantitative Megathread



Hope it helps.
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Hello Bunuel,

Are we always choosing the wider range as the answer?

For -1/9 < x ≤ 2/7 and 2/7 ≤ x < 1,
then we choose -1/9 < x < 1?
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What range of values of x will satisfy the inequality |2x + 02 Sep 2024, 07:46
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lnyngayan
Bunuel
johnnymbikes
What is wrong with the following in the negative scenario B below?

|2x+3| > |7x-2|

(A)
2x + 3 > 7x -2
2x - 7x > -5
-5x > -5
x < -5 / -5
x < 1

(B)
-(2x+3) > 7x-2
2x + 3 < -1 (7x-2)
2x + 3 < -7x + 2
2x + 7x < 2 - 3
9x < -1
x < -1/9 ?

This doesn't match ans choice C

What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

A. x < -1/9 or x > 5

B. -1 < x < 1/9

C. -1/9 < x < 1

D. -1/9 < x < 5

E. x < -1/9 or x > 1

The critical points (also known as transition points or key points) for |2x + 3| and |7x - 2| are x = -3/2 and x = 2/7, respectively. These critical points are where the expressions inside the modulus equal zero, signifying a transition from negative to positive values (or vice-versa).

Considering the two critical points, we analyze three ranges:

1. If x < -3/2, then 2x + 3 < 0 and 7x - 2 < 0. Thus, |2x + 3| = -(2x + 3) and |7x - 2| = -(7x - 2). Hence, for this range we'd get: -(2x + 3) > -(7x - 2), which gives x > 1. Discard this range because it contradicts with the range we consider: x < -3/2.

2. If -3/2 ≤ x ≤ 2/7, then 2x + 3 ≥ 0 and 7x - 2 ≤ 0. Thus, |2x + 3| = 2x + 3 and |7x - 2| = -(7x - 2). Hence, for this range we'd get: 2x + 3 > -(7x - 2), which gives x > -1/9. Combining with the range we consider, we'd get -1/9 < x ≤ 2/7.

3. If x > 2/7, then 2x + 3 > 0 and 7x - 2 > 0. Thus, |2x + 3| = 2x + 3 and |7x - 2| = 7x - 2. Hence, for this range we'd get: 2x + 3 > 7x - 2, which gives x < 1. Combining with the range we consider, we'd get 2/7 ≤ x < 1.

So, we get two ranges: -1/9 < x ≤ 2/7 and 2/7 ≤ x < 1, which give: -1/9 < x < 1.

Answer: C.

10. Absolute Value



For more check Ultimate GMAT Quantitative Megathread



Hope it helps.
­
Hello Bunuel,

Are we always choosing the wider range as the answer?

For -1/9 < x ≤ 2/7 and 2/7 ≤ x < 1,
then we choose -1/9 < x < 1?
Show more
­

We found that |2x + 3| > |7x - 2| is true for all x such that -1/9 < x ≤ 2/7 and 2/7 ≤ x < 1:

------(-1/9)------------(2/7)-------(1)------

------(-1/9)--------------(2/7)------(1)------
Notice that these two ranges form one continuous range: -1/9 < x < 1:

------(-1/9)------------(2/7)------(1)------
Does this make sense?­
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What range of values of x will satisfy the inequality |2x + 27 Nov 2025, 12:20
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Hi Bunuel,

What if the sign is >, then we take the range of greater value always? Like you have mentioned < sign indicates the roots must be between the true values
Bunuel


Since both sides of the inequality are non-negative we can safely square:

\(4x^2+12x+9>49x^2-28x+4\);

\(45x^2-40x-5<0\);

\(9x^2-8x-1<0\);

\((x+\frac{1}{9})(x-1)<0\).

The "roots" are -1/9 and 1 (Solving Quadratic Inequalities: https://gmatclub.com/forum/solving-quad ... 70528.html). "<" sign indicates that the solution must be between the roots: \(-\frac{1}{9}<x<1\).

Answer: C.
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