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</math> if <math>a<0</math>.) | ||
+ | ==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>. | ||
+ | |||
+ | <math>\text{Case 2}: x+3 > 0, x - 1 \leq 0</math> | ||
+ | |||
+ | In this case we have that <math> -3 < x \leq 1</math>. Thus, <math>x+3 +(x-1) = x+1 \implies x = -1</math>. | ||
− | == | + | <math>\text{Case 3}: x+3 \leq 0, x - 1 > 0</math> |
+ | |||
+ | There are obviously no solutions here since <math>x \leq -3</math> and <math>x > 1</math> is a contradiction. | ||
+ | |||
+ | <math>\text{Case 4}: x+3 \leq 0, x - 1 \leq 0</math> | ||
+ | |||
+ | In this case we have <math>x \leq -3</math>. Thus, <math>-x-3+x-1 = x+1 \implies x = -5</math>. | ||
+ | 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 00:07, 30 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 |