8.7 Applications  (Page 2/4)

Person A, working alone, can pour a concrete walkway in 6 hours. Person B, working alone, can pour the same walkway in 4 hours. How long will it take both people to pour the concrete walkway working together?

Step 1:  Let $x$ = the number of hours to pour the concrete walkway working together (since this is what we’re looking for).

Step 2:  If person A can complete the job in 6 hours, A can complete $\frac{1}{6}$ of the job in 1 hour.
If person B can complete the job in 4 hours, B can complete $\frac{1}{4}$ of the job in 1 hour.
If A and B, working together, can complete the job in $x$ hours, they can complete $\frac{1}{x}$ of the job in 1 hour. Putting these three facts into equation form, we have

$\frac{1}{6}+\frac{1}{4}=\frac{1}{x}$
$\begin{array}{l}\begin{array}{ccccccc}\text{Step\hspace{0.17em}3:}\hfill & \hfill & \hfill \frac{1}{6}+\frac{1}{4}& =\hfill & \frac{1}{x}.\hfill & \hfill & \begin{array}{l}\text{An\hspace{0.17em}excluded\hspace{0.17em}value\hspace{0.17em}is\hspace{0.17em}0}\text{.\hspace{0.17em}}\hfill \\ \text{The\hspace{0.17em}LCD\hspace{0.17em}is\hspace{0.17em}12}x\text{.\hspace{0.17em}Multiply\hspace{0.17em}each\hspace{0.17em}term\hspace{0.17em}by\hspace{0.17em}12}x\text{.}\hfill \end{array}\hfill \\ \hfill & \hfill & 12x·\frac{1}{6}+12x·\frac{1}{4}\hfill & =\hfill & 12x·\frac{1}{x}\hfill & \hfill & \hfill \\ \hfill & \hfill & \hfill 2x+3x& =\hfill & 12\hfill & \hfill & \begin{array}{l}\text{Solve\hspace{0.17em}this\hspace{0.17em}nonfractional\hspace{0.17em}}\hfill \\ \text{equation\hspace{0.17em}to\hspace{0.17em}obtain\hspace{0.17em}the\hspace{0.17em}potential\hspace{0.17em}solutions}\text{.}\hfill \end{array}\hfill \\ \hfill & \hfill & \hfill 5x& =\hfill & 12\hfill & \hfill & \hfill \\ \hfill & \hfill & \hfill x& =\hfill & \frac{12}{5}\text{or}x=2\frac{2}{5}\hfill & \hfill & \text{Check\hspace{0.17em}this\hspace{0.17em}potential\hspace{0.17em}solution}\text{.}\hfill \\ \text{Step\hspace{0.17em}4:}\hfill & \hfill & \hfill \frac{1}{6}+\frac{1}{4}& =\hfill & \frac{1}{x}\hfill & \hfill & \hfill \\ \hfill & \hfill & \hfill \frac{1}{6}+\frac{1}{4}& =\hfill & \frac{\frac{1}{12}}{5}.\hfill & \hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ \hfill & \hfill & \hfill \frac{1}{6}+\frac{1}{4}& =\hfill & \frac{5}{12}\hfill & \hfill & \text{The\hspace{0.17em}LCD\hspace{0.17em}is\hspace{0.17em}12.}\text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ \hfill & \hfill & \hfill \frac{2}{12}+\frac{3}{12}& =\hfill & \frac{5}{12}\hfill & \hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ \hfill & \hfill & \hfill \frac{5}{12}& =\hfill & \frac{5}{12}\hfill & \hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \end{array}\hfill \\ \text{Step\hspace{0.17em}5:}\text{Working\hspace{0.17em}together,\hspace{0.17em}A\hspace{0.17em}and\hspace{0.17em}B\hspace{0.17em}can\hspace{0.17em}pour\hspace{0.17em}the\hspace{0.17em}concrete\hspace{0.17em}walkway\hspace{0.17em}in\hspace{0.17em}2}\frac{2}{5}\text{\hspace{0.17em}hours}\text{.}\hfill \end{array}$

An inlet pipe can fill a water tank in 12 hours. An outlet pipe can drain the tank in 20 hours. If both pipes are open, how long will it take to fill the tank?

Step 1:  Let $x$ = the number of hours required to fill the tank.

Step 2:  If the inlet pipe can fill the tank in 12 hours, it can fill $\frac{1}{12}$ of the tank in 1 hour.
If the outlet pipe can drain the tank in 20 hours, it can drain $\frac{1}{20}$ of the tank in 1 hour.
If both pipes are open, it takes $x$ hours to fill the tank. So $\frac{1}{x}$ of the tank will be filled in 1 hour.
Since water is being added (inlet pipe) and subtracted (outlet pipe) we get

$\frac{1}{12}-\frac{1}{20}=\frac{1}{x}$
$\begin{array}{l}\begin{array}{ccccccc}\text{Step\hspace{0.17em}3:}\hfill & \hfill & \hfill \frac{1}{12}-\frac{1}{20}& =\hfill & \frac{1}{x}.\hfill & \hfill & \begin{array}{l}\text{An\hspace{0.17em}excluded\hspace{0.17em}value\hspace{0.17em}is\hspace{0.17em}0}\text{.\hspace{0.17em}The\hspace{0.17em}LCD\hspace{0.17em}is\hspace{0.17em}60}x\text{.\hspace{0.17em}}\hfill \\ \text{Multiply\hspace{0.17em}each\hspace{0.17em}term\hspace{0.17em}by\hspace{0.17em}60}x\text{.}\hfill \end{array}\hfill \\ \hfill & \hfill & \hfill 60x·\frac{1}{12}-60x·\frac{1}{20}& =\hfill & 60x·\frac{1}{x}\hfill & \hfill & \hfill \\ \hfill & \hfill & \hfill 5x-3x& =\hfill & 60\hfill & \hfill & \begin{array}{l}\text{Solve\hspace{0.17em}this\hspace{0.17em}nonfractional\hspace{0.17em}equation\hspace{0.17em}to\hspace{0.17em}obtain\hspace{0.17em}}\hfill \\ \text{the\hspace{0.17em}potential\hspace{0.17em}solutions}\text{.}\hfill \end{array}\hfill \\ \hfill & \hfill & \hfill 2x& =\hfill & 60\hfill & \hfill & \hfill \\ \hfill & \hfill & \hfill x& =\hfill & 30\hfill & \hfill & \text{Check\hspace{0.17em}this\hspace{0.17em}potential\hspace{0.17em}solution}\text{.}\hfill \\ \text{Step\hspace{0.17em}4:}\hfill & \hfill & \hfill \frac{1}{12}-\frac{1}{20}& =\hfill & \frac{1}{x}\hfill & \hfill & \hfill \\ \hfill & \hfill & \hfill \frac{1}{12}-\frac{1}{20}& =\hfill & \frac{1}{30}.\hfill & \hfill & \text{The\hspace{0.17em}LCD\hspace{0.17em}is\hspace{0.17em}60}\text{.}\text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ \hfill & \hfill & \hfill \frac{5}{60}-\frac{3}{60}& =\hfill & \frac{1}{30}\hfill & \hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ \hfill & \hfill & \hfill \frac{1}{30}& =\hfill & \frac{1}{30}\hfill & \hfill & \text{Yes,\hspace{0.17em}this\hspace{0.17em}is\hspace{0.17em}correct.}\hfill \end{array}\hfill \\ \text{Step\hspace{0.17em}5:}\text{With\hspace{0.17em}both\hspace{0.17em}pipes\hspace{0.17em}open,\hspace{0.17em}it\hspace{0.17em}will\hspace{0.17em}take\hspace{0.17em}30\hspace{0.17em}hours\hspace{0.17em}to\hspace{0.17em}fill\hspace{0.17em}the\hspace{0.17em}water\hspace{0.17em}tank}\text{.}\hfill \end{array}$

It takes person A 3 hours longer than person B to complete a certain job. Working together, both can complete the job in 2 hours. How long does it take each person to complete the job working alone?

Step 1:  Let $x$ = time required for B to complete the job working alone. Then, $\left(x+3\right)=$ time required for A to complete the job working alone.
$\begin{array}{l}\begin{array}{ccccccc}\text{Step\hspace{0.17em}2:}\hfill & \hfill & \hfill \frac{1}{x}+\frac{1}{x+3}& =\hfill & \frac{1}{2}.\hfill & \hfill & \hfill \\ \text{Step\hspace{0.17em}3:}\hfill & \hfill & \hfill \frac{1}{x}+\frac{1}{x+3}& =\hfill & \frac{1}{2}.\hfill & \hfill & \begin{array}{l}\text{The\hspace{0.17em}two\hspace{0.17em}excluded\hspace{0.17em}values\hspace{0.17em}are\hspace{0.17em}0\hspace{0.17em}and\hspace{0.17em}}-3.\hfill \\ \text{The\hspace{0.17em}LCD\hspace{0.17em}is\hspace{0.17em}}2x\left(x+3\right)\text{.\hspace{0.17em}}\hfill \end{array}\hfill \\ \hfill & \hfill & \hfill 2x\left(x+3\right)·\frac{1}{x}+2x\left(x+3\right)·\frac{1}{x+3}& =\hfill & 2x\left(x+3\right)·\frac{1}{2}\hfill & \hfill & \hfill \\ \hfill & \hfill & \hfill 2\left(x+3\right)+2x& =\hfill & x\left(x+3\right)\hfill & \hfill & \hfill \\ \hfill & \hfill & \hfill 2x+6+2x& =\hfill & x^2+3x\hfill & \hfill & \begin{array}{l}\text{This\hspace{0.17em}is\hspace{0.17em}a\hspace{0.17em}quadratic\hspace{0.17em}equation\hspace{0.17em}that\hspace{0.17em}can\hspace{0.17em}}\hfill \\ \text{be\hspace{0.17em}solved\hspace{0.17em}using\hspace{0.17em}the\hspace{0.17em}zero-factor\hspace{0.17em}property}\text{.}\hfill \end{array}\hfill \\ \hfill & \hfill & \hfill 4x+6& =\hfill & x^2+3x\hfill & \hfill & \hfill \\ \hfill & \hfill & \hfill x^2-x-6& =\hfill & 0\hfill & \hfill & \hfill \\ \hfill & \hfill & \hfill \left(x-3\right)\left(x+2\right)& =\hfill & 0\hfill & \hfill & \hfill \\ \hfill & \hfill & \hfill x& =\hfill & 3,-2\hfill & \hfill & \text{Check\hspace{0.17em}these\hspace{0.17em}potential\hspace{0.17em}solutions}\text{.}\hfill \end{array}\hfill \\ \text{Step\hspace{0.17em}4:}\text{If\hspace{0.17em}}x=-2,\text{\hspace{0.17em}the\hspace{0.17em}equation\hspace{0.17em}checks,\hspace{0.17em}but\hspace{0.17em}does\hspace{0.17em}not\hspace{0.17em}even\hspace{0.17em}make\hspace{0.17em}physical\hspace{0.17em}sense}\text{.}\hfill \\ \text{If\hspace{0.17em}}x-3\text{,\hspace{0.17em}the\hspace{0.17em}equation\hspace{0.17em}checks}\text{.}\hfill \\ x=3\text{and}x+3=6\hfill \\ \text{Step\hspace{0.17em}5:}\text{Person\hspace{0.17em}B\hspace{0.17em}can\hspace{0.17em}do\hspace{0.17em}the\hspace{0.17em}job\hspace{0.17em}in\hspace{0.17em}3\hspace{0.17em}hours\hspace{0.17em}and\hspace{0.17em}person\hspace{0.17em}A\hspace{0.17em}can\hspace{0.17em}do\hspace{0.17em}the\hspace{0.17em}job\hspace{0.17em}in\hspace{0.17em}6\hspace{0.17em}hours}\text{.}\hfill \end{array}$