How do I solve the following absolute value equation?

Question

I'm having trouble solving this equation:

$$|x+1| = |2x-2|$$

For $x+1 = 2x-2$ and $-(x+1) = -(2x-2)$ I received $x = 3$

and for $-(x+1) = 2x-2$ and $x+1 = -(2x-2)$ I received $x = 1/3$

I tried plugging both into the original equation but they don't always equal.

Draw a picture. The number $x$ is twice as far from $-1$ than it is from $1$. — André Nicolas, Jun 11 '14 at 01:16
Plugging in $x=3$ gives $|3+1|=|2\cdot 3-2| = |4| = 4$ and plugging in $x=1/3$ gives $|1/3 + 1| = |4/3| = |-4/3| = |2/3 - 2|$, so both $x=3$ and $x=1/3$ satisfy the equation. — Nicholas Stull, Jun 11 '14 at 01:26

score 0 · Accepted Answer · answered Jun 11 '14 at 01:25

0

Both solutions work out, for 3: $$ | 3 + 1 | = |2(3) - 2 | $$ $$ | 4 | = | 6 - 2 | $$ $$ 4 = 4 $$

For 1/3:

$$|1/3+1| = |2(1/3) - 2 | $$ $$ 4/3 = | 2/3 - 2 | $$ $$4/3 = | 2/3 - 6/3 | $$ $$4/3 = 4/3$$

answered Jun 11 '14 at 01:25

Dair

1 Answers1