Difference between revisions of "1973 Canadian MO Problems/Problem 2"

Latest revision as of 00:07, 30 December 2015

Find all real numbers that satisfy the equation $|x+3|-|x-1|=x+1$ . (Note: $|a| = a$ if $a\ge 0; |a|=-a$ if $a<0$ .)

We can break this up into cases based upon if $x+3$ and $x-1$ are positive or negative.

$\text{Case 1}: x+3 > 0, x - 1 > 0$

In this case $x > 1$ . Then we have $x+3-x+1 = x+1 \implies x = 3$ .

$\text{Case 2}: x+3 > 0, x - 1 \leq 0$

In this case we have that $-3 < x \leq 1$ . Thus, $x+3 +(x-1) = x+1 \implies x = -1$ .

$\text{Case 3}: x+3 \leq 0, x - 1 > 0$

There are obviously no solutions here since $x \leq -3$ and $x > 1$ is a contradiction.

$\text{Case 4}: x+3 \leq 0, x - 1 \leq 0$

In this case we have $x \leq -3$ . Thus, $-x-3+x-1 = x+1 \implies x = -5$ .

Thus all solutions to this are $x = 3, x= -1,$ and $x = -5.$

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 ==Solution==
-<math>|x+3|-|x-1|=x+1</math>.
 We can break this up into cases based upon if <math>x+3</math> and <math>x-1</math> are positive or negative.

1973 Canadian MO (Problems)
Preceded by Problem 1	1 • 2 • 3 • 4 • 5	Followed by Problem 3