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

Revision as of 00:06, 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 $a<0$ .)

$|x+3|-|x-1|=x+1$ .

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 $x = -5.$

@@ Line 3: / Line 3: @@
 ==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.
+<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==

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