| If x and y satisfy the equations | x | + x + y = 18 and x + | y | − y = 14, then the value of (x − y) is: |
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Answer and solution
Correct answer: C
Solution
Answer: C — 12
3. Establish the sign of y. If y ≥ 0, the second equation gives x = 14. The first then gives y = −10, another contradiction. Therefore y < 0.
5. From the first, y = 18 − 2x. Substitute into the second: x − 2(18 − 2x) = 14, so 5x = 50 and x = 10. Hence y = −2.
6. Compute x − y = 10 − (−2) = 12, option C.
7. Check both original equations: 10 + 10 − 2 = 18 and 10 + 2 − (−2) = 14. Both hold. Proving the signs before removing the modulus prevents lost or invalid branches.
Apply: x≤0 would force y=18 and then x=14, impossible. y≥0 also forces x=14 and fails. Hence x>0, y<0: solve 2x+y=18 and x−2y=14 → x−y=12.
Check: Substitute x=10, y=−2 into both original equations: 10+10−2=18 and 10+2−(−2)=14. The requested subtraction is 10−(−2)=12; it is not the sum 8.
| 1. Write the original equations: | x | + x + y = 18 and x + | y | − y = 14. |
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| 2. Establish the sign of x. If x ≤ 0, then | x | + x = 0, so y = 18. The second equation then gives x = 14, a contradiction. Therefore x > 0. |
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3. Establish the sign of y. If y ≥ 0, the second equation gives x = 14. The first then gives y = −10, another contradiction. Therefore y < 0.
| 4. Substitute | x | = x and | y | = −y. The equations become 2x + y = 18 and x − 2y = 14. |
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5. From the first, y = 18 − 2x. Substitute into the second: x − 2(18 − 2x) = 14, so 5x = 50 and x = 10. Hence y = −2.
6. Compute x − y = 10 − (−2) = 12, option C.
7. Check both original equations: 10 + 10 − 2 = 18 and 10 + 2 − (−2) = 14. Both hold. Proving the signs before removing the modulus prevents lost or invalid branches.
G Strategy shortcut
Spot: Algebra – Modulus and Equations.Apply: x≤0 would force y=18 and then x=14, impossible. y≥0 also forces x=14 and fails. Hence x>0, y<0: solve 2x+y=18 and x−2y=14 → x−y=12.
Check: Substitute x=10, y=−2 into both original equations: 10+10−2=18 and 10+2−(−2)=14. The requested subtraction is 10−(−2)=12; it is not the sum 8.
G Strategy: Spot: Algebra – Modulus and Equations. Apply: x≤0 would force y=18 and then x=14, impossible. y≥0 also forces x=14 and fails. Hence x>0, y<0: solve 2x+y=18 and x−2y=14 → x−y=12. Check: Substitute x=10, y=−2 into both original equations: 10+10−2=18 and 10+2−(−2)=14. The requested subtraction is 10−(−2)=12; it is not the sum 8.