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If |x+5|=3 and |2 y-1| / 3=5, then |x+y|
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09 Mar 2026, 03:04
09 Mar 2026, 03:04
Expert Reply
00:00
Question Stats:
80% (02:21) correct
20% (01:48) wrong based on 15 sessions
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If \(|x+5|=3\) and \(\frac{|2 y-1| }{ 3}=5\), then \(|x+y|\) could equal each of the following EXCEPT
A. 0
B. 6
C. 8
D. 9
E. 15
A. 0
B. 6
C. 8
D. 9
E. 15
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If |x+5|=3 and |2 y-1| / 3=5, then |x+y|
[#permalink]
15 Mar 2026, 23:25
15 Mar 2026, 23:25
1
Expert Reply
Explanation
Solve the absolute value equations first.
1. Solve \(∣x+5∣=3\)
\(x+5=3\) or \(x+5=−3\)
So \(x=−2\) or \(x=−8\)
2. Solve \(\frac{∣2y−1∣}{3}=5\)
Multiply by 3:
\(∣2y−1∣=15\)
\(2y−1=15\) or \(2y−1=−15\)
So, \(y=8\) or \(y=−7\)
3. Compute possible values of \(∣x+y∣\)
Possible (x,y) pairs:
\(x=−2, y=8\), \(∣x+y∣=∣−2+8∣=6\)
\(x=−2,y=−7\), \(∣x+y∣=∣−9∣=9\)
\(x=−8,y=8\), \(∣x+y∣=∣0∣=0\)
\(x=−8,y=−7\), \(∣x+y∣=∣−15∣=15\)
Possible values are:
0, 6, 9, 15
This is an EXCEPT Question so answer is C.
Answer: C
Solve the absolute value equations first.
1. Solve \(∣x+5∣=3\)
\(x+5=3\) or \(x+5=−3\)
So \(x=−2\) or \(x=−8\)
2. Solve \(\frac{∣2y−1∣}{3}=5\)
Multiply by 3:
\(∣2y−1∣=15\)
\(2y−1=15\) or \(2y−1=−15\)
So, \(y=8\) or \(y=−7\)
3. Compute possible values of \(∣x+y∣\)
Possible (x,y) pairs:
\(x=−2, y=8\), \(∣x+y∣=∣−2+8∣=6\)
\(x=−2,y=−7\), \(∣x+y∣=∣−9∣=9\)
\(x=−8,y=8\), \(∣x+y∣=∣0∣=0\)
\(x=−8,y=−7\), \(∣x+y∣=∣−15∣=15\)
Possible values are:
0, 6, 9, 15
This is an EXCEPT Question so answer is C.
Answer: C
gmatclubot
If |x+5|=3 and |2 y-1| / 3=5, then |x+y| [#permalink]
15 Mar 2026, 23:25
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