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Bunuel
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If |x + 5| = 3 and |2y - 1|/3 = 5, then |x + y| could equal each of th 20 Jul 2018, 00:13
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C
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35% (medium)

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75% (02:11) correct 25% (02:20) wrong based on 558 sessions

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If \(|x + 5| = 3\) and \(\frac{|2y - 1|}{3} = 5\), then |x + y| could equal each of the following EXCEPT

A. 0
B. 6
C. 8
D. 9
E. 15
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C
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If |x + 5| = 3 and |2y - 1|/3 = 5, then |x + y| could equal each of th 20 Jul 2018, 00:25
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Bunuel
If \(|x + 5| = 3\) and \(\frac{|2y - 1|}{3} = 5\), then |x + y| could equal each of the following EXCEPT

A. 0
B. 6
C. 8
D. 9
E. 15
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Case 1:

|x+5| = 3

X= -3 or -8


Case 2 :

|2y-1| = 15

y = 8 or -7

Combination of the sum of the x and y:

|(-2) + 8| = 6

| -2 + (-7)| = 9

| -8 + 8| = 0

|-8 +(-7)| = 15

The best answer is C.
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If |x + 5| = 3 and |2y - 1|/3 = 5, then |x + y| could equal each of th 20 Jul 2018, 01:02
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Bunuel
If \(|x + 5| = 3\) and \(\frac{|2y - 1|}{3} = 5\), then |x + y| could equal each of the following EXCEPT
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|x + 5| = 3, solving we get:
x = -2, -8

|2y - 1|/3 = 5
|2y - 1| = 15, solving we get:
y = 8, -7

Combing each values of x and y , we get :
|-2+8| = 6
|-2-7| = 9
|-8+8| = 0
|-8-7| = 15, everything but option C

Hence, C
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If |x + 5| = 3 and |2y - 1|/3 = 5, then |x + y| could equal each of th 30 Oct 2022, 09:28
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Given that |x + 5| = 3 and \(\frac{|2y - 1|}{3} = 5\) and we need to find possible values of |x + y|

|x + 5| = 3
=> x + 5 = 3 or x + 5 = -3 (Watch this video to know about the Basics of Absolute Value)
=> x = 3 - 5 = -2 or x = -3 -5 = -8

\(\frac{|2y - 1|}{3} = 5\)
=> |2y - 1| = 5*3 = 15
=> 2y - 1 = 15 or 2y -1 = -15
=> 2y = 15 + 1 or 2y = -15 + 1
=> 2y = 16 or 2y = -14
=> y = 8 or -7

Possible value of |x + y| are
    1. x = -2, y = 8 => |x + y| = | -2 + 8| = |6| = 6
    2. x = -2, y = -7 => |x + y| = | -2 + -7| = |-9| = 9
    3. x = -8, y = 8 => |x + y| = | -8 + 8| = |0| = 0
    4. x = -8, y = -7 => |x + y| = | -8 + -7| = |-15| = 15

So, Answer will be C
Hope it helps!

Watch the following video to learn How to Solve Absolute Value Problems

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If |x + 5| = 3 and |2y - 1|/3 = 5, then |x + y| could equal each of th 19 Mar 2026, 09:17
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Deconstructing the Question

We need the possible values of \(|x+y|\), given

\(|x+5|=3\)

and

\(\frac{|2y-1|}{3}=5\)

The fastest approach is to find all possible values of x and y, then combine them.

Step-by-step

From

\(|x+5|=3\)

we get

\(x+5=3\) or \(x+5=-3\)

So

\(x=-2\) or \(x=-8\)

From

\(\frac{|2y-1|}{3}=5\)

multiply both sides by \(3\):

\(|2y-1|=15\)

So

\(2y-1=15\) or \(2y-1=-15\)

Thus

\(2y=16 \Rightarrow y=8\)

or

\(2y=-14 \Rightarrow y=-7\)

Now combine the cases.

If \(x=-2\) and \(y=8\), then \(x+y=6\), so \(|x+y|=6\)

If \(x=-2\) and \(y=-7\), then \(x+y=-9\), so \(|x+y|=9\)

If \(x=-8\) and \(y=8\), then \(x+y=0\), so \(|x+y|=0\)

If \(x=-8\) and \(y=-7\), then \(x+y=-15\), so \(|x+y|=15\)

So the possible values are

\(0,\ 6,\ 9,\ 15\)

The only value that cannot occur is

\(8\)

Answer C
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