Chapter 7.4, Problem 6P

To determine

To find: The original fraction as mentioned in the problem.

Answer to Problem 6P

The original fraction is $\frac{3}{4}$.

Explanation of Solution

Given information:

The numerator of a fraction is $1$ less than denominator.

And numerator and denominator is both increased by $4$ .

Formula used:

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

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

We will use cross multiplication method.

Calculation:

Consider that the denominator is $x$ then numerator is $x-1$ .

Then the original fraction is $\frac{x-1}{x}$ .

Now the numerator and denominator is increased by $4$ .

Then the fraction is $\frac{x+3}{x+4}=\frac{x-1}{x}+\frac{1}{8}\Rightarrow \frac{x+3}{x+4}-\frac{x-1}{x}=\frac{1}{8}$ .

Recall that we will take the L.C.M of the denominator $\left(x+4,x\right)$ is $x\left(x+4\right)$ .

Therefore $\frac{x+3}{x+4}-\frac{x-1}{x}=\frac{1}{8}\Rightarrow \frac{x*\left(x+3\right)-\left(x-1\right)*\left(x+4\right)}{x\left(x+4\right)}=\frac{1}{8}$

Therefore on further solving we have

$\begin{array}{l}\frac{x*\left(x+3\right)-\left(x-1\right)*\left(x+4\right)}{x\left(x+4\right)}=\frac{1}{8}\Rightarrow \frac{x^2+3x-x^2-3x+4}{x\left(x+4\right)}=\frac{1}{8}\\ \Rightarrow \frac{4}{x^2+4x}=\frac{1}{8}\end{array}$

Recall that we will use cross multiplication method.

  $\frac{4}{x^2+4x}=\frac{1}{8}\Rightarrow \left(8\right)*4=\left(x^2+4x\right)\Rightarrow 32=x^2+4x\Rightarrow x^2+4x-32=0$ .

Recall that we will use quadratic equation $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

Where $a=1,b=4,c=-32$ .

Therefore

$\begin{array}{l}\frac{-\left(4\right)\pm \sqrt{{\left(4\right)}^2+4*1*32}}{2*1}\Rightarrow \frac{-4\pm \sqrt{16+128}}{2}\Rightarrow \frac{-4\pm \sqrt{144}}{2}\\ \Rightarrow \frac{-4\pm 12}{2}\Rightarrow \frac{-4+12}{2},\frac{-4-12}{2}\Rightarrow \frac{8}{2},\frac{-16}{2}\Rightarrow 4,-8\end{array}$ .

Therefore the original fraction is $\frac{4-1}{4}$ = $\frac{3}{4}$ .