Given: $|x| + x + y = 15$ ... (equation 1) $x + |y| - y = 20$ ... (equation 2) Understanding Absolute Values: Remember that absolute value removes the negative sign: If a number is positive or zero: $|a| = a$ If a number is negative: $|a| = -a$ Since we don't know if $x$ and $y$ are positive or negative, we need to consider all possibilities. This is because the absolute value expressions $|x|$ and $|y|$ behave differently depending on the sign of the variable.

Smart Observation Before Starting Cases: From equation 1: $|x| + x = 15 - y$ From equation 2: $|y| - y = 20 - x$ Key Insight: If $x \geq 0$, then $|x| + x = x + x = 2x$ If $x < 0$, then $|x| + x = (-x) + x = 0$ If $y \geq 0$, then $|y| - y = y - y = 0$ If $y < 0$, then $|y| - y = (-y) - y = -2y$

Case 1: $x \geq 0$ and $y \geq 0$ Using our insights: $|x| + x = 2x$ and $|y| - y = 0$ Our equations become: $2x = 15 - y$ ... (A) $0 = 20 - x$ $\rightarrow x = 20$ ... (B) Substituting $x = 20$ into equation (A): $2(20) = 15 - y$ $40 = 15 - y$ $y = -25$ Check: We assumed $y \geq 0$, but got $y = -25 < 0$. This contradicts our assumption. Conclusion: Case 1 is impossible.

Case 2: $x \geq 0$ and $y < 0$ Using our insights: $|x| + x = 2x$ and $|y| - y = -2y$ Our equations become: $2x = 15 - y$ ... (A) $-2y = 20 - x$ ... (B) From equation (B): $x = 20 + 2y$ Substituting into equation (A): $2(20 + 2y) = 15 - y$ $40 + 4y = 15 - y$ $5y = -25$ $y = -5$ Finding $x$: $x = 20 + 2(-5) = 20 - 10 = 10$ Check: $x = 10 \geq 0$ and $y = -5 < 0$ Conclusion: Case 2 gives us a valid solution: $x = 10, y = -5$

Case 3: $x < 0$ and $y \geq 0$ Using our insights: $|x| + x = 0$ and $|y| - y = 0$ Our equations become: $0 = 15 - y$ $\rightarrow y = 15$ $0 = 20 - x$ $\rightarrow x = 20$ Check: We assumed $x < 0$, but got $x = 20 > 0$. This contradicts our assumption. Conclusion: Case 3 is impossible.

Case 4: $x < 0$ and $y < 0$ Using our insights: $|x| + x = 0$ and $|y| - y = -2y$ Our equations become: $0 = 15 - y$ $\rightarrow y = 15$ $-2y = 20 - x$ From the first equation: $y = 15$ From the second equation: $-2(15) = 20 - x$ $-30 = 20 - x$ $x = 50$ Check: We assumed both $x < 0$ and $y < 0$, but got $x = 50 > 0$ and $y = 15 > 0$. This contradicts our assumptions. Conclusion: Case 4 is impossible.

Final Answer: Only Case 2 gives us a valid solution: $x = 10$ and $y = -5$ Therefore: $x - y = 10 - (-5) = 10 + 5 = 15$

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