Question

In the following exercises, solve.

$1-\frac{7}{q}=\frac{-6}{q^2}$

Short answer

The solution of the equation $1-\frac{7}{q}=\frac{-6}{q^2}$is

$q=1\&6$

Step-by-step solution

Step 1. Given information 

The given equation is

$1-\frac{7}{q}=\frac{-6}{q^2}$

Step 2. Clear the fraction  

Some terms of the equation are not defined for the$q=0$

sorole="math" localid="1647790931408" $1-\frac{7}{q}=\frac{-6}{q^2},q\ne 0$

The least common denominator of all denominators in the equation is$q^2$

multiply both sides of the equation by the least common denominator $q^2$

$$\begin{gathered} q^2\left(1-\frac{7}{q}\right)=q^2\left(\frac{-6}{q^2}\right) \\ {q}^2\left(1\right)-q^2\left(\frac{7}{q}\right)=q^2\left(\frac{-6}{q^2}\right) \\ {q}^2-7q=-6 \\ {q}^2-7q+6=0 \end{gathered}$$

Step 3. Solution of equation  

Factorize the equation

$$\begin{gathered} q^2-7q+6=0 \\ {q}^2-6q-q+6=0 \\ q(q-6)-1(q-6)=0 \\ \left(q-1\right)\left(q-6\right)=0 \end{gathered}$$

Use the Zero Product Property

$$\begin{gathered} q-1=0 \\ q=1 \end{gathered}$$

or$$\begin{gathered} q-6=0 \\ q=6 \end{gathered}$$

So solutions of the equation arerole="math" localid="1647791077334" $q=1\&6$

Step 4. Verification  

Substitute $q=1$in the equation

$$\begin{gathered} 1-\frac{7}{q}=\frac{-6}{q^2} \\ 1-\frac{7}{1}\stackrel{?}{=}\frac{-6}{1^2} \\ 1-7\stackrel{?}{=}-6 \\ -6=-6 \end{gathered}$$

The left-hand side is equal to the right-hand side so the solution$q=1$ is correct.

Substitute $q=6$in the equation

$$\begin{gathered} 1-\frac{7}{6}=\frac{-6}{q^2} \\ \frac{6}{6}-\frac{7}{6}\stackrel{?}{=}\frac{-6}{6^2} \\ \frac{-1}{6}\stackrel{?}{=}\frac{-6}{36} \\ \frac{-1}{6}=\frac{-1}{6} \end{gathered}$$

The left-hand side is equal to the right-hand side so the solution $q=6$is correct.