# How do you solve and check for extraneous solutions in abs(2t-3) = t?

Aug 1, 2015

$\textcolor{red}{t = 3}$ is a solution.
$\textcolor{red}{t = 1}$ is an extraneous solution.

SOLVE

$| 2 t - 3 | = t$

We need to write two different equations without the absolute value symbols and solve for $t$.

These equations are:

(1): $\left(2 t - 3\right) = t$
(2): $- \left(2 t - 3\right) = t$

Solve Equation 1:

$2 t - 3 = t$

Subtract $t$ from each side.

$t - 3 = 0$

Add $3$ to each side.

$t = 3$

Solve Equation 2:

−(2t-3) = t

Remove parentheses.

$- 2 t + 3 = t$

Add $2 t$ to each side.

$3 = 3 t$

Divide each side by $3$.

$t = 1$

The solutions are $t = 1$ and $t = 3$.

CHECK FOR EXTRANEOUS SOLUTIONS:

If $t = 1$,

$| 2 t - 3 | = t$
$| 2 \left(1\right) - 3 | = 3$
$| 2 - 3 | = 5$
$| - 1 | = 5$
$1 = 5$

This is impossible, so $t = 1$ is an extraneous solution.

If $t = 3$,

$| 2 t - 3 | = t$
$| 2 \left(3\right) - 3 | = 3$
$| 6 - 3 | = 3$
$| 3 | = 3$
$3 = 3$

$t = 3$ is a solution.