Solve each equation. Check your solution.

$\text{3}\left|2t-1\right|=15$

On real number line, the distance of a number from 0 is the absolute value. It is always nonnegative.

The absolute value of a function is expressed as, if x is a real number then absolute value of xis defined as.

$\left|x\right|=x$ when x is greater than or equal to 0 . In other words, the absolute value of x is x when x is either positive or zero.

$\left|x\right|=-x$ when x is less than 0 . In other words, the absolute value of x is opposite of x when x is negative.

The absolute value of 5 is expressed as $\left|5\right|=5$ and of $-5$ is expressed as $\left|-5\right|=5$

Consider the provided equation.

$3\left|2t-1\right|=15$

Divide both sides by 3.

$\left|2t-1\right|=5$

Recall the concept of absolute value and apply it.

Case 1. $2t-1=5$

Add 1 on both the sides.

Divide both sides by 2.

$t=3$

Case 2. $2t-1=-5$

Add 1 on both the sides.

Divide both sides by 2.

$t=-2$

Therefore, there are two possible solutions for the equation $3\left|2t-1\right|=15$ that are $t=3$ and $t=-2$.

Substitute the value $t=3$ in the equation $3\left|2t-1\right|=15$.

Simplify it further as.

$15=15$

Since, this is true so the value $t=3$ satisfy the equation $3\left|2t-1\right|=15$.

Substitute the value $t=-2$ in the equation $3\left|2t-1\right|=15$.

Simplify it further as.

$15=15$

Since, this is true so the value $t=-2$ satisfy the equation $3\left|2t-1\right|=15$.

Thus, the solution set is $\left\{-2,3\right\}$.

Hence, the solutions of the equation are $t=-2$ and $t=3$.