Difference between revisions of "1973 Canadian MO Problems/Problem 2"
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==Problem== | ==Problem== | ||
| − | Find all real numbers that satisfy the equation <math>|x+3|-|x-1|=x+1</math>. (Note: <math>|a| = a</math> if <math>a\ge 0; |a|=-a if a<0</math>.) | + | Find all real numbers that satisfy the equation <math>|x+3|-|x-1|=x+1</math>. (Note: <math>|a| = a</math> if <math>a\ge 0; |a|=-a</math> if <math>a<0</math>.) |
==Solution== | ==Solution== | ||
| + | We can break this up into cases based upon if <math>x+3</math> and <math>x-1</math> are positive or negative. | ||
| + | |||
| + | <math>\text{Case 1}: x+3 > 0, x - 1 > 0</math> | ||
| + | |||
| + | In this case <math>x > 1</math>. Then we have <math>x+3-x+1 = x+1 \implies x = 3</math>. | ||
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| + | <math>\text{Case 2}: x+3 > 0, x - 1 \leq 0</math> | ||
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| + | In this case we have that <math> -3 < x \leq 1</math>. Thus, <math>x+3 +(x-1) = x+1 \implies x = -1</math>. | ||
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| + | <math>\text{Case 3}: x+3 \leq 0, x - 1 > 0</math> | ||
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| + | There are obviously no solutions here since <math>x \leq -3</math> and <math>x > 1</math> is a contradiction. | ||
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| + | <math>\text{Case 4}: x+3 \leq 0, x - 1 \leq 0</math> | ||
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| + | In this case we have <math>x \leq -3</math>. Thus, <math>-x-3+x-1 = x+1 \implies x = -5</math>. | ||
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| + | Thus all solutions to this are <math>x = 3, x= -1,</math> and <math>x = -5.</math> | ||
==See also== | ==See also== | ||
Latest revision as of 23:07, 29 December 2015
Problem
Find all real numbers that satisfy the equation 
. (Note: 
if 
if 
.)
Solution
We can break this up into cases based upon if 
and 
are positive or negative.
In this case 
. Then we have 
.
In this case we have that 
. Thus, 
.
There are obviously no solutions here since 
and is a contradiction.
In this case we have . Thus, 
.
Thus all solutions to this are 
and 
See also
| 1973 Canadian MO (Problems) | ||
| Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 | Followed by Problem 3 |
