Refer to the workbook on pages 76 and 77. There problems are set up and worked out here for you to follow.
Sarah has \($400\) in her savings account and she has to pay \($15\) each month to her parents for her cell phone. Darius has \($50\) and he saves \($20\) each month from his job walking dogs for his neighbor. At this rate, when will Sarah and Darius have the same amount of money? How much money will they each have?
| Let \(x\) be the number of months and \(y\) be the total amount of money. | |||
| Equation #1: | \(y=-15x+400\) | Equation #2: | \(y=20x+50\) |
| Use the Substitution Method: | \(\begin{array}{l} y=\color{blue}{-15x+400} & \color{blue}{y}=20x+50 \\ \implies \color{blue}{-15x+400}=20x+50 \\ \implies 350=35x \\ \implies 10=x \\ y=-15\color{red}{x}+400 & y=-15(\color{red}{10})+400 \\ \implies y=-150+400 \\ \implies y=250 \\ (10,250)\;\color{green}{\checkmark} \end{array}\) | ||
| It will take \(10\) months for Sarah and Darius to have \($250\). | |||
The admission fee at a local zoo is \($1.50\) for children and \($4.00\) for adults. On a certain day, \(2,200\) people enter the zoo and \($5,050\) is collected. How many children and how many adults attended the zoo that day?
| Let \(c\) denote the number of children and \(a\) the number of adults. | |
| Equations: | \(c+a=2200\) and \(1.50c+4.00a=5050\) |
| Use the Elimination Method: | Multiply first equation by \(-1.50\): \(-1.50c-1.50a=-3300\) |
| \(\begin{array}{l} -1.50c-1.50a=-3300 \\ \underline{+1.50c+4.00a=5050} \\ \hfill 2.50a=1750 \\ \implies a=700 \\ \implies c+700=2200 \\ \implies c=1500 \\ \implies (700,1500)\;\color{green}{\checkmark}\end{array}\) | |
| There were \(1,500\) children and \(700\) adults in attendance. | |
Castel's school is selling tickets to a play. On the first day of ticket sales, the school sold 10 adult tickets and 6 student tickets for a total of \($126\). The school took in \($72\) on the second day by selling 1 adult ticket and 6 student tickets. What is the price of one adult ticket and one student ticket?
| Let \(x\) be the price of an adult ticket and \(y\) the price of a student ticket. | |
| Equations: | \(10x+6y=126\) and \(x+6y=72\) |
| Elimination works best here since \(6y\) is in both equations. | Multiply the second equation by \(-1\): \(-1\cdot(x+6y=72)\implies -1x-6y=-72\) |
| \(\begin{array}{l} \hfill{} 10x+6y=126 \\ +\underline{-1x-6y=-72} \\ \implies \hfill 9x=54 \\ \implies \large{\frac{9x}{9}}\normalsize{=}\large{\frac{54}{9}} \\ \implies \normalsize{x=6} \end{array}\) | |
| Just for practice, use the Elimination Method again to find the \(y\)-value. | Multiply the second equation by \(-10\): \(-10\cdot(x+6y=72)\implies -10x-60y=-720\) |
| \(\begin{array}{l} \hspace{22pt} 10x+6y=126 \\ \underline{+ -10x-60y=-720} \\ \implies \hfill -54y=-594 \\ \implies \large{\frac{-54y}{-54}}\normalsize{=}\large{\frac{-594}{-54}} \\ \implies \normalsize{y=11} \\ \implies (6,11)\;\color{green}{\checkmark} \end{array}\) | |
| The cost of one adult ticket is \($6\) and the cost of a student ticket is \($11\). | |