What range of values of x will satisfy the inequality |2x +

Question

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

Bunuel · Accepted Answer

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.

amit2k9 · Answer

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.

gauravsoni · Answer

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.

Bunuel · Answer

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

gauravsoni · Answer

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

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.

Bunuel · Answer

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. (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.

ashsim · Answer

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)?

WoundedTiger · Answer

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

NGGMAT · Answer

hi bunnel

what do u mean when u say the following:

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??

Bunuel · Answer

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.

himanshujovi · Answer

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.

Bunuel · Answer

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

Hope it helps.

Vinayprajapati · Answer

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.

johnnymbikes · Answer

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

Bunuel · Answer

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 Theory Math: Absolute value (Modulus) Absolute Value: Tips and hints Properties of Absolute Values on the GMAT Absolute modulus : A better understanding GMAT Question Patterns: Abolute Value The Reason Behind Absolute Value Questions on the GMAT Questions Inequality and absolute value questions from my collection The E-GMAT Question Series on ABSOLUTE VALUE DS Questions PS Questions For more check Ultimate GMAT Quantitative Megathread Hope it helps.

chetanraghav1705 · Answer

Here is my explanation to it. Hope you like it.

Posted from my mobile device

lnyngayan · Answer

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?

Bunuel · Answer

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?

Azeem123 · Answer

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

What range of values of x will satisfy the inequality |2x +

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What range of values of x will satisfy the inequality |2x +

Prep Toolkit

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