If |8x − 7| 3x + 8, then each of the following could be true EXCEPT:

Question

If  | 8x − 7 | > 3x + 8 , then each of the following could be true EXCEPT 1 < x < 3

B.  − 1 < x < 0

C.  2 < x < 4

D.  x = 3.5

E.  x = − 1

CrackverbalGMAT · Accepted Answer

Opening the modulus, we have -(3x + 8) > 8x - 7 > 3x + 8 Solving for Right Hand side 8x - 7 > 3x + 8 8x - 3x > 15 5x > 15 x > 3 Solving for Left Hand side 8x - 7 < -(3x + 8) 8x + 3x < -8 + 7 11x < -1 x < -1/11 or 0.09 The same is depicted on the number line below. All Options other than A are in the shaded region. Option A Arun Kumar

chetan2u · Answer

This is a very simple MOD question so follow the proper method. | 8x − 7 | > 3x + 8 Two cases... I) Critical point 1 : x>\frac{7}{8} 8x-7>0....................8x-7>3x+8..........................x>3 II) Critical point 2 : x\leq{\frac{7}{8}} 8x-7<0................................7-8x>3x+8............................11x<-1......................x<\frac{-1}{11} The range that will give NO is -1/11\leq{x}\leq{3} . Any choice completely inside this range is the answer let us look for overlap in each A. 1 < x < 3 ....YES B. − 1 < x < 0 .... -1 3*2 + 8.......7>14 ...NO, and our answer B. − 1 < x < 0 ......let x=-0.5..... | 8*-0.5 − 7 | > 3*-0.5+ 8.......-11>6.5 ...YES C. 2 < x < 4 ......let x=3..... | 8*3 − 7 | > 3*3+ 8.......17>17 ...NO....But if x=3.5, the answer is YES One can answer even C as the answer if one is NOT careful to check all choices.

Basshead · Answer

8x - 7 > 3x + 8 5x > 15 x > 3 8x - 7 < -(3x + 8) 8x - 7 < -3x - 8 11x < -1 x < -1/11 Therefore, x cannot be 1

Naomilobo16 · Answer

Can someone please tell me what am i doing wrong here

-(8x-7) < 3x+8
-8x+7<3x+8
-1<11x
-1/11<x

pearljiandani · Answer

The question is |8x-7| > 3x+8, your sign is wrong.

BrushMyQuant · Answer

Given that | 8x − 7 | > 3x + 8 and we need to find the range of values of x to solve the problem Let's solve the problem using two method Method 1: Substitution | 8x − 7 | > 3x + 8 let's substitute values in the range of each answer choice and check if it satisfies the equation A. 1 < x < 3 Let's take x = 2 and substitute in | 8x − 7 | > 3x + 8 => | 8*2 − 7 | > 3*2 + 8 => | 16 - 7| > 6 + 8 => | 9 | > 14 Which is FALSE. So, NOT POSSIBLE B. − 1 < x < 0 Let's take x = -0.5 and substitute in | 8x − 7 | > 3x + 8 => | 8*(-0.5) − 7 | > 3*(-0.5) + 8 => | -4 - 7| > -1.5 + 8 => | -11 | > 6.5 => 11 > 6.5 Which is TRUE. So, POSSIBLE C. 2 < x < 4 Let's take x = 3.5 and substitute in | 8x − 7 | > 3x + 8 => | 8*3.5 − 7 | > 3*3.5 + 8 => | 28 - 7| > 10.5 + 8 => | 21 | > 18.5 => 21 > 18.5 Which is TRUE. So, POSSIBLE D. x = 3.5 . same as option C E. x = − 1 Let's take x = -1 and substitute in | 8x − 7 | > 3x + 8 => | 8*(-1) − 7 | > 3*(-1) + 8 => | -8 - 7| > -3 + 8 => | -15 | > 5 => 15 > 55 Which is TRUE. So, POSSIBLE Method 2: Algebra To solve this we need to open |8x - 7| and we can do that by taking two cases To solve this we need to open |8x - 7| and we can do that by taking two cases - Case 1: 8x - 7 >= 0 => x >= \frac{7}{8} ( ~0.875) Since, 8x - 7 >= 0 => | 8x - 7 | = 8x - 7 => 8x - 7 > 3x + 8 => 8x - 3x > 8 + 7 => 5x > 15 => x > \frac{15}{5} => x > 3 Intersection of x >= 7/8 and x > 3 is x > 3 - Case 2: 8x - 7 < 0 => x < \frac{7}{8} ( ~0.875) Since, 8x - 7 < 0 => | 8x - 7 | = -(8x - 7) = -8x + 7 => -8x + 7 > 3x + 8 => -8x - 3x > 8 - 7 => -11x > 1 => x < \frac{-1}{11} (= -0.09) Intersection of x < 7/8 and x < \frac{-1}{11} is x < \frac{-1}{11} So, x < \frac{-1}{11} or x > 3 is the solution range Let's look at the answer choices now A. 1 < x < 3 => NOT POSSIBLE , as it is not in the range of x < \frac{-1}{11} or x > 3 is the solution range B. − 1 < x < 0 => Could be true as x < \frac{-1}{11} and part of − 1 < x < 0 lies in this range C. 2 < x < 4 => Could be true as x > 3 and part of 2 < x < 4 lies in this range D. x = 3.5 => TRUE as x > 3 is one range E. x = − 1 => TRUE as x< \frac{-1}{11} is one range So, Answer will be A Hope it helps! Watch the following video to learn the Basics of Absolute Values https://www.youtube.com/watch?v=huFiBk8PaGY

Mehakgyl · Answer

Why did flip the sign in the 2nd scenario?

BrushMyQuant · Answer

Check the Method 2 which i explained above in my solution. Its easier to take that as

-(8x - 7) > 3x + 8 as the second case and solve.
 Hope it helps!

RahulPaul · Answer

Why wouldn't case 2 open with -8x+7>3x +8 and give x<-1/5 instead of x<-1/11? What am I doing wrong?

Bunuel · Answer

-8x + 7 > 3x + 8
-11x > 1

So:
x < -1/11

If |8x − 7| > 3x + 8, then each of the following could be true EXCEPT:

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To solve this we need to open \|8x - 7\| and we can do that by taking two cases
-Case 1: 8x - 7 >= 0 => x >= \(\frac{7}{8}\) ( ~0.875) Since, 8x - 7 >= 0 => \| 8x - 7 \| = 8x - 7 => 8x - 7 > 3x + 8 => 8x - 3x > 8 + 7 => 5x > 15 => x > \(\frac{15}{5}\) => x > 3 Intersection of x >= 7/8 and x > 3 is x > 3	-Case 2: 8x - 7 < 0 => x < \(\frac{7}{8}\) ( ~0.875) Since, 8x - 7 < 0 => \| 8x - 7 \| = -(8x - 7) = -8x + 7 => -8x + 7 > 3x + 8 => -8x - 3x > 8 - 7 => -11x > 1 => x < \(\frac{-1}{11}\) (= -0.09) Intersection of x < 7/8 and x < \(\frac{-1}{11}\) is x < \(\frac{-1}{11}\)

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