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How many different values of x are there such that |x + 10| = 2x + 8? 13 May 2024, 01:30
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

55% (hard)

Question Stats:

61% (01:20) correct 39% (01:11) wrong based on 622 sessions

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­How many different values of \(x\) are there such that \(|x+10|=2x+8\)?

A. None
B. 1
C. 2
D. 3
E. Infinitely many­

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How many different values of x are there such that |x + 10| = 2x + 8? 01 Jun 2024, 07:15
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Official Solution:

How many different values of \(x\) are there such that \(|x+10|=2x+8\)?

A. None
B. 1
C. 2
D. 3
E. Infinitely many


The left-hand side of the equation is an absolute value, which is always non-negative. Thus, the right-hand side must also be non-negative: \(2x+8 \ge 0\), which gives us \(x \ge -4\). For this range, \(x+10\) is positive, so \(|x+10|=x+10\). Therefore, \(|x+10|=2x+8\) can be rewritten as \(x+10=2x+8\). Solving for \(x\) gives \(x=2\). Therefore, only one value of \(x\) satisfies \(|x+10|=2x+8\).


Answer: B­
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How many different values of x are there such that |x + 10| = 2x + 8? 13 May 2024, 02:10
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Bunuel
­How many different values of \(x\) are there such that \(|x+10|=2x+8\)?

A. None
B. 1
C. 2
D. 3
E. Infinitely many­


 
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­
\(|x+10|=2x+8\)

As the RHS of the equation is modulus, \(2x + 8 \geq 0\)

⇒ \(x \geq -4\)

\(|x+10|=2x+8\)

Squaring both sides we get

\(x^2 + 100 + 20x = 4x^2 + 64 + 32x\)

\(3x^2 + 12x - 36 = 0\)

\(x^2 + 4x - 12 = 0\)

\(x + 6x - 2x - 12 = 0\)

\((x + 6)(x - 2) = 0\)

Roots = \(-6, +2\)

However, the value of \(x\) must be greater than \(-4.\) Hence, only one value, \(+2\), holds in this case.

Option B
 ­
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How many different values of x are there such that |x + 10| = 2x + 8? 13 May 2024, 02:32
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Bunuel
­How many different values of \(x\) are there such that \(|x+10|=2x+8\)?

A. None
B. 1
C. 2
D. 3
E. Infinitely many­


­
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­
after opening modulus x + 10 = 2x + 8 --- (1)
and - x - 10 = 2x + 8 --- (2) 

after solving for x we will get x = 2 and -6

Hence C.­
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How many different values of x are there such that |x + 10| = 2x + 8? 13 May 2024, 04:29
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\(­|x + 10| = 2x + 8\)

Making the aboslute positive: 

\(­x + 10 = 2x + 8\)

\(x = 2\) 

Plugging it back into the original equation holds as both side come to \(12\).

Making the aboslute negative: 

\(-­x - 10 = 2x + 8\)

\(3x = -18\)

\(x = -6\)

Plugging this back into the original equation does not hold as the left side will be \(4\) and the right side will be \(-4\)

Therefore only \(2\) works.

ANSWER B

 
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How many different values of x are there such that |x + 10| = 2x + 8? Updated on: 18 Nov 2024, 10:18
Originally posted by saswat1987 on 27 May 2024, 10:16.
Last edited by saswat1987 on 18 Nov 2024, 10:18, edited 1 time in total.
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­The solution provided is 100% correct, but the problem why people get such questions wrong (I was also in the same league once :roll:) is that many of us do not put back the value of x (2,-6) into the original equation and verify which one of them satisfies the equation.

Those of you who have put back the value will mark answer as 1 and rest will mark as 2 (out of excitement or ignorance).

P.S. -- The key takeaway is that in Absolute Algebra questions always verify the solutions found, as some solutions will satisfy the equation and some will not.­­

edit: corrected the spelling and made the answer explanation clear and concise.
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How many different values of x are there such that |x + 10| = 2x + 8? 17 Nov 2024, 04:56
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|x+10|=2x+8 , so we will have two cases
-Case 1: x + 10 ≥ 0, => x ≥ -10

=> |x + 10| = x + 10
=> x + 10 = 2x + 8
=> x = 2


But the condition was x ≥ -10 and 2 ≥ -10
=> x = 2 is a SOLUTION		-Case 2: x + 10 ≤ 0 => x ≤ -10

|x + 10| = -(x + 10) = -x - 10
=> -x - 10 = 2x + 8
=> 3x = -18
=> x = \(\frac{-18}{3}\) = -6

But the condition was x ≤ -10 and -6 IS NOT ≤ -10
=> x = -6 is NOT a SOLUTION

=> There is 1 possible value of x i.e. x = 2

So, Answer will be B
Hope it helps!

Watch the following video to MASTER Absolute Values

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How many different values of x are there such that |x + 10| = 2x + 8? 26 May 2025, 05:55
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Why do we have to check if x >-4, why not just determine roots and apply to check if x+10=2x+8 Bunuel
gmatophobia
Bunuel
­How many different values of \(x\) are there such that \(|x+10|=2x+8\)?

A. None
B. 1
C. 2
D. 3
E. Infinitely many­


­
\(|x+10|=2x+8\)

As the RHS of the equation is modulus, \(2x + 8 \geq 0\)

⇒ \(x \geq -4\)

\(|x+10|=2x+8\)

Squaring both sides we get

\(x^2 + 100 + 20x = 4x^2 + 64 + 32x\)

\(3x^2 + 12x - 36 = 0\)

\(x^2 + 4x - 12 = 0\)

\(x + 6x - 2x - 12 = 0\)

\((x + 6)(x - 2) = 0\)

Roots = \(-6, +2\)

However, the value of \(x\) must be greater than \(-4.\) Hence, only one value, \(+2\), holds in this case.

Option B
­
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How many different values of x are there such that |x + 10| = 2x + 8? 26 May 2025, 09:34
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Why do we have to check if x >-4, why not just determine roots and apply to check if x+10=2x+8 Bunuel
gmatophobia
Bunuel
­How many different values of \(x\) are there such that \(|x+10|=2x+8\)?

A. None
B. 1
C. 2
D. 3
E. Infinitely many­


­
\(|x+10|=2x+8\)

As the RHS of the equation is modulus, \(2x + 8 \geq 0\)

⇒ \(x \geq -4\)

\(|x+10|=2x+8\)

Squaring both sides we get

\(x^2 + 100 + 20x = 4x^2 + 64 + 32x\)

\(3x^2 + 12x - 36 = 0\)

\(x^2 + 4x - 12 = 0\)

\(x + 6x - 2x - 12 = 0\)

\((x + 6)(x - 2) = 0\)

Roots = \(-6, +2\)

However, the value of \(x\) must be greater than \(-4.\) Hence, only one value, \(+2\), holds in this case.

Option B
­
Show more


The reason we check x ≥ -4 is because the left side of the equation is |x + 10|, which is always non-negative, so the right side 2x + 8 must also be non-negative. That gives us the restriction x ≥ -4. For x ≥ -4, x + 10 is positive, so |x + 10| equals x + 10. This lets us directly solve x + 10 = 2x + 8, which simplifies the solution and avoids the conventional methods.
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