Chapter 7.4, Problem 33WE

To determine

To solve: the expression as mentioned in the problem.

Answer to Problem 33WE

The solution exist and it is $n=0,3$.

Explanation of Solution

Given information:

The expression is $\frac{n-2}{n}-\frac{n-3}{n-6}=\frac{1}{n}$.

Formula used:

We will use cross multiplication method.

We will take the variable term to left hand side and constant term to right hand side.

We will use L.C.M of the denominator.

We will use the quadratic equation formula $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

Calculation:

Consider the expression $\frac{n-2}{n}-\frac{n-3}{n-6}=\frac{1}{n}$.

Here variable is $n$ in the mentioned expression above.

Recall that we will use L.C.M of the denominator.

Therefore the L.C.M of $\left(n,n-6\right)$ is $n\left(n-6\right)$.

Therefore $\frac{n-2}{n}-\frac{n-3}{n-6}=\frac{1}{n}\Rightarrow \frac{\left(n-2\right)*\left(n-6\right)-\left(n\right)*\left(n-3\right)}{n\left(n-6\right)}=\frac{1}{n}\Rightarrow \frac{n^2-8n+12-n^2+3n}{n^2-6n}=\frac{1}{n}\Rightarrow \frac{-5n+12}{n^2-6n}=\frac{1}{n}$.

Recall that we will use cross multiplication method.

  $\frac{12-5n}{n^2-6n}=\frac{1}{n}\Rightarrow n*\left(12-5n\right)=1*\left(n^2-6n\right)\Rightarrow 12n-5n^2=n^2-6n$.

Recall that we will take the variable term to left hand side and constant term to right hand side.

Therefore $12n-5n^2=n^2-6n\Rightarrow -6n^2+18n=0\Rightarrow -6n\left(n-3\right)=0\Rightarrow n=0,3$.

Therefore the value of the expression has a solution $n=0,3$.