Q53. - Algebra 1

Question

Solve each equation. Then graph the solution set.

$\left|\frac{1}{3}d-6\right|=15$

Step-by-Step Solution

Answer

The Solution set is $\{63,-27\}$.

Step 1. Case when the modulus expression is positive .

The absolute value or modulus of a real number x, denoted $\left|x\right|$, it is the non-negative value of $x$ without regard to its sign.

$\left|x\right|=x$ if $x$ is positive,

$\left|x\right|=-x$ if $x$ is negative,

$\left|0\right|=0$.

For example, the absolute value of 5 is 5, and the absolute value of −5 is also 5.

 

We are given the expression $\left|\frac{1}{3}d-6\right|=15$ assuming a positive value for the modulus expression we get:

$\begin{array}{l}\frac{1}{3}d-6=15\\ \frac{1}{3}d=15+6\\ \frac{1}{3}d=21\\ d=21\left(3\right)\\ d=63\end{array}$

Thus $d=63$ is one of the solutions.

Step 2. Case when the modulus expression is negative.

$\begin{array}{l}\frac{1}{3}d-6=-15\\ \frac{1}{3}d=-15+6\\ \frac{1}{3}d=-9\\ d=-9\left(3\right)\\ d=-27\end{array}$ 

Thus $d=-27$ is one of the solutions.

Step 3. Graphing the solution set.

The solution set is $\{63,-27\}$, we graph these values on the number line as shown below: