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 \(x+y\).

|x| + y = 10 & x + |y| = 6