Q68. - Algebra 2

Question

Solve each equation. Check your solution. 

$68.\left|4r+3\right|-7=10$

Step-by-Step Solution

Answer

The solutions of the equation are $r=-5$ and $r=\frac{7}{2}$.

Step-1 – Apply the concept of Absolute value

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 x is 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.

Step-2 –Example of Absolute value

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

Step-3 – Simplify the equation

Consider the provided equation.

$\left|4r+3\right|-7=10$

Add 7 to both the sides.

$\begin{array}{l}|4r+3|-7+7=10+7\\ |4r+3|=17\end{array}$

Recall the concept of absolute value and apply it.

Case 1. $4r+3=17$

Subtract 3 from both the sides.

$\begin{array}{l}4r+3-3=17-3\\ 4r=14\end{array}$

Subtract 3 from both the sides.

$\begin{array}{c}4r+3-3=17-3\\ 4r=14\end{array}$

Divide both sides by 4.

$\begin{array}{c}r=\frac{14}{4}\\ r=\frac{7}{2}\end{array}$

Case 2. $4r+3=-17$

Subtract 3 from both the sides.

$\begin{array}{c}4r+3-3=-17-3\\ 4r=-20\end{array}$

Divide both sides by 4.

$\begin{array}{c}r=\frac{-20}{4}\\ r=-5\end{array}$

Therefore, there are two possible solutions for the equation $\left|4r+3\right|-7=10$ that are $r=-5$ and $r=\frac{7}{2}$.

Step-4 – Verify the solutions

Substitute the value $r=\frac{7}{2}$ in the equation $\left|4r+3\right|-7=10$.

$\begin{array}{c}|4\left(\frac{7}{2}\right)+3|-7=10\\ |14+3|-7=10\\ \left|17\right|-7=10\\ 17-7=10\end{array}$

Simplify it further as.

$10=10$

Since, this is true so the value $r=\frac{7}{2}$satisfy the equation $\left|4r+3\right|-7=10$.

Substitute the value $r=-5$ in the equation $\left|4r+3\right|-7=10$.

$\begin{array}{c}|4(-5)+3|-7=10\\ |-20+3|-7=10\\ |-17|-7=10\\ 17-7=10\end{array}$

Simplify it further as.

$10=10$

Since, this is true so the value $r=-5$ satisfy the equation $\left|4r+3\right|-7=10$.

Thus, the solution set is $\left\{-5,\frac{7}{2}\right\}$.

Hence, the solutions of the equation are $r=-5$ and $r=\frac{7}{2}$.