8.7 Applications

When the same number is added to the numerator and denominator of the fraction $\frac{3}{5},$ the result is $\frac{7}{9}.$ What is the number that is added?

Step 1: Let  $x=$ the number being added. $$\begin{array}{lllllll}\text{Step\hspace{0.17em}2:}\hfill & & \hfill \frac{3+x}{5+x}& =& \frac{7}{9}.\hfill & & \\ \text{Step\hspace{0.17em}3:}\hfill & & \hfill \frac{3+x}{5+x}& =& \frac{7}{9}.\hfill & & \text{An\hspace{0.17em}excluded\hspace{0.17em}value\hspace{0.17em}is\hspace{0.17em}}-\text{5}\text{.}\\ & & & & & & \text{The\hspace{0.17em}LCD\hspace{0.17em}is\hspace{0.17em}9}\left(5+x\right)\text{.\hspace{0.17em}}\\ & & & & & & \text{Multiply\hspace{0.17em}each\hspace{0.17em}term\hspace{0.17em}by\hspace{0.17em}9}\left(5+x\right).\\ & & \hfill 9\left(5+x\right)·\frac{3+x}{5+x}& =& 9\left(5+x\right)·\frac{7}{9}\hfill & & \\ & & \hfill 9\left(3+x\right)& =& 7\left(5+x\right)\hfill & & \\ & & \hfill 27+9x& =& 35+7x\hfill & & \\ & & \hfill 2x& =& 8\hfill & & \\ & & \hfill x& =& 4\hfill & & \text{Check\hspace{0.17em}this\hspace{0.17em}potential\hspace{0.17em}solution}\text{.}\\ \text{Step\hspace{0.17em}4:}\hfill & & \hfill \frac{3+4}{5+4}& =& \frac{7}{9}.\hfill & & \text{Yes,\hspace{0.17em}this\hspace{0.17em}is\hspace{0.17em}correct.}\\ \text{Step\hspace{0.17em}5:}\hfill & & \text{The\hspace{0.17em}number\hspace{0.17em}added\hspace{0.17em}is\hspace{0.17em}4}\text{.}& & & & \end{array}$$

Two thirds of a number added to the reciprocal of the number yields $\frac{25}{6}.$ What is the number?

Step 1:    Let $x$ = the number.

Step 2:    Recall that the reciprocal of a number $x$ is the number $\frac{1}{x}$ .

       $\frac{2}{3}·x+\frac{1}{x}=\frac{25}{6}$

$\begin{array}{lllllll}\text{Step 3:}\hfill & \hfill & \hfill \frac{2}{3}·x+\frac{1}{x}& =\hfill & \frac{25}{6}\hfill & \hfill & \text{The LCD is}6x.\text{Multiply each term by}6x.\hfill \\ \hfill & \hfill & 6x·\frac{2}{3}x+6x·\frac{1}{x}\hfill & =\hfill & 6x·\frac{25}{6}\hfill & \hfill & \hfill \\ \hfill & \hfill & \hfill 4x^2+6& =\hfill & 25x\hfill & \hfill & \begin{array}{l}\text{Solve this nonfractional quadratic equation to obtain the}\hfill \\ \text{potential solutions}\text{. (Use the zero-factor property}\text{.)}\hfill \end{array}\hfill \\ \hfill & \hfill & \hfill 4x^2-25x+6& =\hfill & 0\hfill & \hfill & \hfill \\ \hfill & \hfill & \hfill (4x-1)(x-6)& =\hfill & 0\hfill & \hfill & \hfill \\ \hfill & \hfill & \hfill x=\frac{1}{4},& \hfill & 6\hfill & \hfill & \text{Check these potential solutions}\text{.}\hfill \end{array}$

Step 4:    Substituting into the original equation, it can be that both solutions check.

Step 5:    There are two solutions: $\frac{1}{4}$ and 6.