# Thread: Absolute Value Equations

1. ## Absolute Value Equations

Could someone please tell me how to solve the following without graphing?

(1) $|x+4|+|x-7|=|2x-1|$

(2) $|x-|x-1||=\lfloor x \rfloor$ (Greatest integer function of x)

2. Originally Posted by Winding Function
Could someone please tell me how to solve the following without graphing?

(1) $|x+4|+|x-7|=|2x-1|$

(2) $|x-|x-1||=\lfloor x \rfloor$ (Greatest integer function of x)

For number (1) if you state the boundaries for what the absolute value signs would do to the terms you can solve it as a simple algebraic equation.

For example: |x-7| take away the absolute value signs and state that it is only true for positive real numbers x-7 > or equal to 0

|x+4| is just x+4 because the absolute value sign doesn't affect positive real numbers.

I hope this was helpful

3. Um, the problem didn't specify any boundaries. Could someone please show me how to solve the equations? Thanks!

4. For $x \geq 7$:

$|x + 4| + |x - 7| = |2x - 1|$ is equivalent to

$x + 4 + x - 7 = 2x - 1$
$2x - 3 = 2x - 1$
which is impossible.

For $0.5 \leq x < 7$:

$|x + 4| + |x - 7| = |2x - 1|$ is equivalent to

$x + 4 + 7 - x = 2x - 1$
$11 = 2x - 1$
$12 = 2x$
$x = 6$
and that's one solution.

For $-4 \leq x < 0.5$:

$|x + 4| + |x - 7| = |2x - 1|$ is equivalent to

$x + 4 + 7 - x = 1 - 2x$
$11 = 1 - 2x$
$10 = -2x$
$x = -5$
but this solution is not in the correct domain.

For $x < -4$:

$|x + 4| + |x - 7| = |2x - 1|$ is equivalent to

$-x - 4 + 7 - x = 1 - 2x$
$-2x + 3 = -2x + 1$
which is impossible.

Hence, the one solution to this equation is x = 6.