Chapter 1.6, Problem 70STP

To determine

To solve: the given equation for $y$ . Check the result.

Answer to Problem 70STP

The value of $y$ in the given equation is $y=5\text{or}y=-\frac{35}{3}$ .

Explanation of Solution

Given information:

Given equation

  $\left|3y+1\text{0}\right|=25$

Concept used:

Absolute value of a function a is given by

  $\left|a\right|=\{\begin{array}{l}aa>0\\ -aa<0\\ 0a=0\end{array}$

And

  $\begin{array}{l}\left|-a\right|=a\\ \left|a\right|=a\end{array}$

More over

  $\begin{array}{l}\left|x+a\right|=b\\ \Rightarrow x+a=b\text{or}x+a=-b\end{array}$

Calculation:

Consider the given equation,

  $\left|3y+1\text{0}\right|=25$

Now, solve the equation as shown,

  $\begin{array}{c}\left|3y+1\text{0}\right|=25\\ 3y+1\text{0}=25\text{or}3y+1\text{0}=-25\\ 3y+1\text{0}-1\text{0}=25-1\text{0}\text{or}3y+1\text{0}-1\text{0}=-25-1\text{0}\\ 3y=15\text{or}3y=-35\\ y=\frac{15}{3}\text{or}y=-\frac{35}{3}\\ y=5\text{or}y=-\frac{35}{3}\end{array}$

Now, check the result by back substitution.

  $\begin{array}{c}\left|3\left(5\right)+1\text{0}\right|=25\\ \left|15+1\text{0}\right|=25\\ \left|25\right|=25\\ 25=25\end{array}$

And

  $\begin{array}{c}\left|3\left(-\frac{35}{3}\right)+1\text{0}\right|=25\\ \left|-35+1\text{0}\right|=25\\ \left|-25\right|=25\\ 25=25\end{array}$

Which is true.

Hence, the value of $y$ in the given equation is $y=5\text{or}y=-\frac{35}{3}$ .