How do you solve abs(t+1)=4t +3?

Mar 2, 2018

Appliying definition of absolute value. See details

Explanation:

We define absolute value of a number as

$\left\mid x \right\mid = x$ if $x \ge 0$ and

$\left\mid x \right\mid = - x$ if $x \le 0$

With this in mind, lets aplly to our equation in t

$\left\mid t + 1 \right\mid = t + 1$ if $t + 1 \ge 0$, it say: $t \ge - 1$

$\left\mid t + 1 \right\mid = - \left(t + 1\right)$ if $t + 1 < 0$, it say: $t < - 1$

In the first case ($t \ge - 1$) :

$t + 1 = 4 t + 3$

$1 - 3 = 4 t - t$; thus $t = - \frac{2}{3}$ This value is valid because is bigger than -1

In the second case ($t < - 1$) :

$- \left(t + 1\right) = 4 t + 3$

$- t - 1 = 4 t + 3$

$t = - \frac{4}{5}$ this value is invalid because our initial restriction $t < - 1$ is not verified by $t = - \frac{4}{5}$

Mar 2, 2018

$t = - \frac{2}{3}$

Explanation:

$\text{the value inside the absolute value bars can be }$
$\text{positive or negative}$

$\text{thus there are 2 possible solutions}$

$t + 1 = 4 t + 3 \leftarrow \textcolor{b l u e}{\text{positive inside bars}}$

$\text{subtract "(t+1)" from both sides}$

$\Rightarrow 0 = 3 t + 2$

$\Rightarrow 3 t = - 2 \Rightarrow t = - \frac{2}{3} \leftarrow \textcolor{red}{\text{possible solution}}$

$- t - 1 = 4 t + 3 \leftarrow \textcolor{b l u e}{\text{negative inside bars}}$

rArr5t=-4rArrt=-4/5larrcolor(red)"possible solution "

$\textcolor{b l u e}{\text{As a check}}$

$| - \frac{2}{3} + 1 | = | \frac{1}{3} | = \frac{1}{3} \text{ and } - \frac{8}{3} + \frac{9}{3} = \frac{1}{3}$

$\text{both sides are equal hence "x=-2/3" is a solution}$

$| - \frac{4}{5} + 1 | = \frac{1}{5} \text{ and } - \frac{16}{5} + \frac{15}{5} = - \frac{1}{5}$

$\frac{1}{5} \ne - \frac{1}{5} \text{ hence "t=-4/5" is not a solution}$