|x| + y = 10 & x + |y| = 6
November 3, 2024 2025-11-03 23:29|x| + y = 10 & x + |y| = 6
Solving \( |x|+y=10 \) and \( x+|y|=6 \) — Clean Case Method
We’ll check sign-cases for \(x\) and \(y\), prune contradictions, and get \(x+y\) fast.
Problem
Given: \( |x| + y = 10 \) and \( x + |y| = 6 \). Find \( x + y \).
Absolute value splits by sign, so consider the four sign cases for \((x,y)\):
\((\ge0,\ge0)\), \((\ge0,<0)\), \((<0,\ge0)\), \((<0,<0)\).
Case analysis (step by step)
Case 1: \(x\ge0,\; y\ge0\)
\[
x+y=10,\qquad x+y=6
\]
Contradiction (\(10\neq 6\)). No solution here.
Case 2: \(x\ge0,\; y<0\)
Then \(|x|=x,\ |y|=-y\):
\[
x+y=10,\qquad x-y=6
\]
Add: \(2x=16\Rightarrow x=8\). Substitute to get \(y=2\), which violates \(y<0\).
Hence, no solution.
Case 3: \(x<0,\; y\ge0\)
\(|x|=-x,\ |y|=y\):
\[
-x+y=10,\qquad x+y=6
\]
Subtract second from first: \(-2x=4\Rightarrow x=-2\). Then \(y=8\).
Both satisfy \(x<0\) and \(y\ge0\). ✅
Case 4: \(x<0,\; y<0\)
\(|x|=-x,\ |y|=-y\):
\[
-x+y=10,\qquad x-y=6
\]
Add: \(-2x=16\Rightarrow x=-8\). Then from \(x-y=6\): \(-8-y=6\Rightarrow y=-14\) — but this actually fits \(y<0\).
Check in the first equation: \(-(-8)+(-14)=8-14=-6\neq 10\). So contradiction → discard.
Conclusion
The only valid solution is from Case 3: \(x=-2,\ y=8\). Therefore,
\[
x+y=-2+8=\boxed{6}.
\]
Quick review (remember this)
- Absolute value means branch by sign: \( |x|=\begin{cases} x&(x\ge0)\\ -x&(x<0)\end{cases} \).
- Write equations for each sign case; contradictions kill cases quickly.
- Always verify the final pair in the original equations.
Practice
- Solve \( |x|+|y|=10 \) and \( x-y=2 \). Find \(x,y\).
- Given \( |x|+y=7 \) and \( x+|y|=9 \), find all possible values of \(x+y\).
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