Solve the equation. Check the solutions.

$4|3t+8|=16t$

Absolute value describes the distance from zero that a number is on the number line,without considering direction. The absolute value of a number is never negative.

Consider the expression $4|3t+8|=16t$.

Divide both sides of the equation by 4 as follows:

$\begin{array}{l}4|3t+8|=16t\\ |3t+8|=4t\end{array}$

Apply the absolute rule as follows:

Case 1. When $3t+8=-4t$.

$\begin{array}{l}3t+8=-4t\\ 3t+8+4t=0\\ 7t+8=0\\ 7t=-8\\ t=-\frac{8}{7}\end{array}$

Case 2. When $3t+8=4t$

$\begin{array}{l}3t+8=4t\\ 3t+8-4t=0\\ -t+8=0\\ -t=-8\\ t=8\end{array}$

Therefore, the value is $t=8,-\frac{8}{7}$.

Substitute $t=8,-\frac{8}{7}$in the equation $4|3t+8|=16t$and simplify as follows:

$\begin{array}{l}4|3t+8|=16t\\ 4|3\left(8\right)+8|=16\left(8\right)\\ 4|24+8|=128\\ 4\left|32\right|=128\\ 128=128\end{array}$

$\begin{array}{l}4|3t+8|=16t\\ 4|3(-\frac{8}{7})+8|=16(-\frac{8}{7})\\ 4|-\frac{24}{7}+8|=-\frac{128}{7}\\ 4\left|\frac{-24+56}{7}\right|=-\frac{128}{7}\\ 4\left|\frac{32}{7}\right|=-\frac{128}{7}\\ \frac{128}{7}=-\frac{128}{7}\end{array}$

Since the left-hand side and right-hand side is equal in the first case but not in the second case thus the values are $t=8$.