Question

In the following exercises, solve.

$\frac{1}{r+3}=\frac{4}{2r}$

Short answer

The solution of the equation $\frac{1}{r+3}=\frac{4}{2r}$is

$r=-6$

Step-by-step solution

Step 1. Given information  

The given equation is

$\frac{1}{r+3}=\frac{4}{2r}$

Step 2. Solution of equation    

Terms of the equation are not defined for$r\ne 0\&r\ne -3$

sorole="math" localid="1647800002686" $\frac{1}{r+3}=\frac{4}{2r},r\ne 0,r\ne -3$

The least common denominator of all denominators in the equation is$2r(r+3)$

multiply both sides of the equation by the least common denominator $2r(r+3)$

$$\begin{gathered} 2r(r+3)\left(\frac{1}{r+3}\right)=2r(r+3)\left(\frac{4}{2r}\right) \\ 2r\left(1\right)=(r+3)\left(4\right) \\ 2r=4r+12 \\ r=-6 \end{gathered}$$

So the solution of the equation is$r=-6$

Step 3. Verification 

Substitute $r=-6$in the equation

$$\begin{gathered} \frac{1}{r+3}=\frac{4}{2r} \\ \frac{1}{\left(-6\right)+3}\stackrel{?}{=}\frac{4}{2\left(-6\right)} \\ \frac{1}{-3}\stackrel{?}{=}\frac{4}{-12} \\ \frac{1}{-3}=\frac{1}{-3} \end{gathered}$$

The left-hand side is equal to the right-hand side so the solution is correct.