Question

In the following exercise solve$\frac{z}{12}+\frac{z+3}{3z}=\frac{1}{z}$

Short answer

The solution is$z=-4$

Step-by-step solution

Step 1. Given information

$\frac{z}{12}+\frac{z+3}{3z}=\frac{1}{z}$

Step 2. Solve

The least common denominator of all denominators in the equation is $\left(12\right)\left(3z\right)$

Clear the fraction by multiplying both sides of the equation by the LCD.

$$\begin{gathered} \left(12\right)\left(3z\right)\left(\frac{z}{12}+\frac{z+3}{3z}\right)=\left(12\right)\left(3z\right)\left(\frac{1}{z}\right) \\ \left(12\right)\left(3z\right)\left(\frac{z}{12}\right)+\left(12\right)\left(3z\right)\left(\frac{z+3}{3z}\right)=\left(12\right)\left(3z\right)\left(\frac{1}{z}\right) \\ 3z^2+12\left(z+3\right)=36 \\ 3z^2+12z=0 \\ z+4=0 \\ z=-4 \end{gathered}$$