Chapter 11, Problem 18PT

To determine

Solve the equation $\frac{v}{3}+\frac{v}{v+5}=\frac{-4}{v+5}$ and check the solutions.

Answer to Problem 18PT

Solutions of the equation $\frac{v}{3}+\frac{v}{v+5}=\frac{-4}{v+5}$ are $v=-6\text{and}v=-2$

Explanation of Solution

Given:

The equation: $\frac{v}{3}+\frac{v}{v+5}=\frac{-4}{v+5}$

Concept Used:

Zero-Product Property: If we multiply two (or more) things together and the result is equal to zero, then we know that  at least one  of those things that we multiplied must also have been equal to zero.

Calculation:

Given the Equation: $\frac{v}{3}+\frac{v}{v+5}=\frac{-4}{v+5}$

Solve for the unknown variable ‘ v’ .

Step 1: The LCM of the denominators is $3(v+5)$ . Multiply each term of the equation by the LCM $3(v+5)$ to get rid from the denominators.

  $\begin{array}{l}3(v+5)·\frac{v}{3}+3(v+5)·\frac{v}{v+5}=3(v+5)·\frac{-4}{v+5}\\ v·(v+5)+3v=-12\\ v^2+5v+3v+12=-12+12\\ v^2+8v+12=0\end{array}$

Find the factor form of the equation: $v^2+8v+12=0$

Factor form: $v^2+8v+12=0\Rightarrow (v+6)(v+2)=0$

Now use zero product property:

  $\begin{array}{l}{v}^2+8v+12=0\\ (v+6)(v+2)=0\\ v+6=0\text{and}v+2=0[\text{Zero-Product Property}]\\ v=-6\text{and}v=-2\end{array}$

  $\text{Check}\text{the}\text{solutions}\text{when}v=-6\text{and}v=-2$

  $v^2+8v+12=0v^2+8v+12=0$

  $(-6·-6)+8·(-6)+12=0(-2·-2)+8·(-2)+12=0$

  $36-48+12=04-16+12=0$

  $48-48=0[\text{Checked}]16-16=0[\text{Checked}]$

Thus, Solutions of the equation $\frac{v}{3}+\frac{v}{v+5}=\frac{-4}{v+5}$ are $v=-6\text{and}v=-2$