If x≠4, what is the range of the solutions of the equation |14–x|=24/(

Question

If x≠4, what is the range of the solutions of the equation  |14–x|=\frac{24}{(x−4)} ?

A. 2
B. 6
C. 8
D. 20
E. 32

BrentGMATPrepNow · Accepted Answer

There are 3 steps to solving equations involving ABSOLUTE VALUE:
1. Apply the rule that says:   If |x| = k, then x = k and/or x = -k  
2. Solve the resulting equations
3. Plug solutions into original equation to check for extraneous roots

Given: |14–x|=24/(x−4)

So, we need to check  14–x = 24/(x−4)  and  14–x = -24/(x−4)

14–x = 24/(x−4) 
Multiply both sides by (x-4) to get: (14–x)(x−4) = 24
Expand: -x² + 18x - 56 = 24
Rearrange to get: x² - 18x + 80 = 0
Factor: (x - 10)(x - 8) = 0
So, x = 10 or x = 8

Test each solution: 
If x = 10, then we get: |14–10|=24/(10−4) 
Simplify: |4|=4 PERFECT!
So,   x = 10   is a possible solution

If x = 8, then we get: |14–8|=24/(8−4) 
Simplify: |6|=6 PERFECT!
So,   x = 8   is a possible solution
 
 14–x = -24/(x−4) 
Multiply both sides by (x-4) to get: (14–x)(x−4) = -24
Expand: -x² + 18x - 56 = -24
Rearrange to get: x² - 18x + 32 = 0
Factor: (x - 2)(x - 16) = 0
So, x = 2 or x = 16

Test each solution: 
If x = 2, then we get: |14–2|=24/(2−4) 
Simplify: |12|=-12 DOESN'T WORK
So, x = 2 is NOT a possible solution

If x = 16, then we get: |14–16|=24/(16−4) 
Simplify: |-2|=2 PERFECT!
So,   x = 16   is a possible solution

So, the possible solutions are   8, 10 and 16  
Range =   16   -   8   = 8

Answer:  C

RELATED VIDEO
 https://www.youtube.com/watch?v=LHz74_lvLI8

Kurtosis · Answer

|14 - x| = 24/(x - 4)

(x - 4)(14 - x) = 24
x^2 - 18x + 80 = 0
x = 8 or 10

(x - 4)(14 - x) = -24
x^2 - 18x + 32 = 0
x = 16 or 2 but 2 is not a valid solution since  |14 - 2| 
eq {-12}

Range = 16 - 8 = 8

Answer: C

rekhabishop · Answer

pushpitkc  , Is it possible to solve this question using the number line?

It took me more than 2 minutes to solve this question. I normally prefer the number line, but in this case I just couldn't use it.

GyMrAT · Answer

Given: |14–x|=\frac{24}{(x−4)} For 14 - x > 0, we have x < 14 (14-x)(x-4) = 24, solving we get x = 10 or 8 < 14, hence both solutions are good. For 14 - x < 0, we have x > 14 (x-14)(x-4) = 24. solving we get x = 2 or 16, only x = 16 > 14 is good. hence we have x = 8, 10, 16 Range of Solutions = 16 - 8 = 8 Answer C. Thanks, GyM

adam613 · Answer

There are 3 steps to solving equations involving ABSOLUTE VALUE:
1. Apply the rule that says:   If |x| = k, then x = k and/or x = -k  
2. Solve the resulting equations
3. Plug solutions into original equation to check for extraneous roots

Given: |14–x|=24/(x−4)

So, we need to check  14–x = 24/(x−4)  and  14–x = -24/(x−4)

14–x = 24/(x−4) 
Multiply both sides by (x-4) to get: (14–x)(x−4) = 24
Expand: -x² + 18x - 56 = 24
Rearrange to get: x² - 18x + 80 = 0
Factor: (x - 10)(x - 8) = 0
So, x = 10 or x = 8

Test each solution: 
If x = 10, then we get: |14–10|=24/(10−4) 
Simplify: |4|=4 PERFECT!
So,   x = 10   is a possible solution

If x = 8, then we get: |14–8|=24/(8−4) 
Simplify: |6|=6 PERFECT!
So,   x = 8   is a possible solution
 
 14–x = -24/(x−4) 
Multiply both sides by (x-4) to get: (14–x)(x−4) = -24
Expand: -x² + 18x - 56 = -24
Rearrange to get: x² - 18x + 32 = 0
Factor: (x - 2)(x - 16) = 0
So, x = 2 or x = 16

Test each solution: 
If x = 2, then we get: |14–2|=24/(2−4) 
Simplify: |12|=-12 DOESN'T WORK
So, x = 2 is NOT a possible solution

If x = 16, then we get: |14–16|=24/(16−4) 
Simplify: |-2|=2 PERFECT!
So,   x = 16   is a possible solution

So, the possible solutions are   8, 10 and 16  
Range =   16   -   8   = 8

Answer:  C

14–x = 24/(x−4)
Multiply both sides by (x-4) to get: (14–x)(x−4) = 24
 Expand: -x² + 18x - 56 = 24
Rearrange to get: x² - 18x + 80 = 0 
Factor: (x - 10)(x - 8) = 0
So, x = 10 or x = 8

When you do -56 - 24 why did you get a positive 80 and not negative 80?

Bambi2021 · Answer

I solved it the following way, but maybe it was just a coincidence that my approach worked in this case:

When we arrive at (x-4)(14-x) = 24, I just thought that the range must be between the values of x when any of the factors is 0.

x = 4 -> 0
x = 14 -> 0

Range is 5-13 = 8.

Guess I was just lucky this time.

MBAB123 · Answer

Interesting point, but I don't think we are told that X is an integer. Just that when we solve, the max value is 16 and the minimum value is 8.

BrushMyQuant · Answer

Given that x≠4 and we need to find the range of the solutions of the equation \(|14–x|=\frac{24}{(x−4)}\)

Now, in order to solve this equation we need to deal with the Absolute Value of |14-x|
We can do that by taking two cases

Case 1: 14 -x >= 0 or x <= 14
=> | 14-x | = 14 -x
=> \(14–x = \frac{24}{(x−4)}\)
=> (14-x) * ( x-4) = 24
We can either use Algebra (open the equation and solve) or can substitute values

Substitution

(14-x) * ( x-4) = 4 * 6
If we put x = 10 then we will get
(14-10) * (10-4) = 4*6
So, x = 10

(14-x) * ( x-4) = 6 * 4
If we put x = 8 then we will get
(14-8) * (8-4) = 6*4
So, x = 8

Algebra

(14-x) * ( x-4) = 24
=> 14x - 56 - \(x^2\) + 4x = 24
=> - \(x^2\) + 18x - 56 - 24 = 0
=> - \(x^2\) + 18x - 80 = 0
Multiply by -1 we get
=> \(x^2\) - 18x + 80 = 0
=> \(x^2\) - 8x - 10x + 80 = 0
=> x*(x-8) - 10*(x-8) = 0
=> (x-8) * (x-10) = 0
=> x = 8 or 10

Both values are <= 14
=> x can be 8 or 10

Case 2: 14 -x < 0 or x > 14
=> | 14-x | = -(14 -x) = x - 14
=> \(x-14 = \frac{24}{(x−4)}\)
=> (x-14) * ( x-4) = 24
We can either use Algebra (open the equation and solve) or can substitute values

Substitution

(x-14) * (x-4) = 2 * 12
If we put x = 16 then we will get
(16-14) * (16-4) = 2*12
So, x = 16
Since 16 > 14
=> x = 16 is a solution

Algebra

(x-14) * ( x-4) = 24
=> \(x^2\) - 4x -14x + 56 = 24
=> \(x^2\) - 18x + 56 - 24 = 0
=> \(x^2\) - 18x + 32 = 0
=> \(x^2\) - 2x - 16x + 32 = 0
=> x*(x-2) - 16*(x-2) = 0
=> (x-2) * (x-16) = 0
=> x = 2 or 16
But x > 14
=> x = 16 is a solution

Combining both the cases possible values of x are 8, 10, 16

Range = Highest value - Lowest value = 16-8 = 8

So, Answer will be C
Hope it helps!

Watch the following video to learn the Basics of Absolute Values

MK010 · Answer

Hi @KarishmaB,

In the 2nd case - Multiply both sides by (x-4) to get: (14–x)(x−4) = -24, upon solving the equation we received the values 16 and 2. However these are not negative values as per our assumption for the expression. Hence, shouldn't they be outright rejected instead of testing?

I am trying to understand the testing rationale here.

KarishmaB · Answer

When we have a mix and x and |x| in an equation, there is only one thing to do - Take cases. |14–x|=\frac{24}{(x−4)} is the same as |x-14|=\frac{24}{(x−4)} Two cases: Case 1: x >= 14 x-14=\frac{24}{(x−4)} x^2 -18x + 32 = 0 x = 2 or 16 Only 16 is greater than 14 hence only 16 is an acceptable answer. Case 2: x< 14 x-14=\frac{-24}{(x−4)} x^2 - 18x + 80 = 0 x = 8 or 10 Both are acceptable since both are less than 14 Range of solutions = Greatest solution - Least solution = 16 - 8 = 8 Answer (C) Check this post: https://anaprep.com/algebra-the-why-beh ... questions/

KarishmaB · Answer

MK010 
First, simplify |14 - x| as |x - 14| since both are the same. 
After that, take two cases as I have shown above. Once you get your solutions, all you need to do is check the original condition and accept solutions as per that. The ones that do not fall under our original condition should be outright rejected. No testing needed.

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If x≠4, what is the range of the solutions of the equation |14–x|=24/(

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If x≠4, what is the range of the solutions of the equation |14–x|=24/(

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