Difference between revisions of "1988 AHSME Problems/Problem 17"
(Created page with "==Problem== If <math>|x| + x + y = 10</math> and <math>x + |y| - y = 12</math>, find <math>x + y</math> <math>\textbf{(A)}\ -2\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ \frac{1...") |
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==Solution== | ==Solution== | ||
| + | We proceed by casework: | ||
| + | Case 1: x, y >= 0 | ||
| + | |||
| + | Thus we have 2x + y = 10, x = 12. This makes y < 0, a contradiction. | ||
| + | |||
| + | Case 2: x >= 0 > y | ||
| + | |||
| + | 2x + y = 10, x - 2y = 12. By multiplying the first equation by two and adding, we have 4x + x = 2*10 + 12 thus x = 32/5. Since the question only specifies 1 solution, plugging in x = 32/5 into the first equation yields y = -14/5 thus x + y = 18/5 <math>fbox{A}</math>. | ||
== See also == | == See also == | ||
Revision as of 15:36, 2 February 2018
Problem
If 
and 
, find 
Solution
We proceed by casework:
Case 1: x, y >= 0
Thus we have 2x + y = 10, x = 12. This makes y < 0, a contradiction.
Case 2: x >= 0 > y
2x + y = 10, x - 2y = 12. By multiplying the first equation by two and adding, we have 4x + x = 2*10 + 12 thus x = 32/5. Since the question only specifies 1 solution, plugging in x = 32/5 into the first equation yields y = -14/5 thus x + y = 18/5 
.
See also
| 1988 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
