11.4 Applications  (Page 4/2)

The sum of two numbers is 37. One number is 5 larger than the other. What are the numbers?

$\begin{array}{llll}\text{Step 1:}\hfill & \text{Let\hspace{0.17em}}x\hfill & =\hfill & \text{smaller\hspace{0.17em}number}.\hfill \\ \hfill & y\hfill & =\hfill & \text{larger\hspace{0.17em}number}.\hfill \end{array}$

Step 2:  There are two relationships.
     (a) The Sum is 37.
         $x+y=37$
     (b) One is 5 larger than the other.
         $y=x+5$
        $\begin{array}{lll}\{\begin{array}{l}x+y=37\\ y=x+5\end{array}\hfill & \hfill & \begin{array}{l}\left(1\right)\\ \left(2\right)\end{array}\hfill \end{array}$

Step 3:   $\begin{array}{lll}\{\begin{array}{l}x+y=37\\ y=x+5\end{array}\hfill & \hfill & \begin{array}{l}\left(1\right)\\ \left(2\right)\end{array}\hfill \end{array}$
 We can easily solve this system by substitution. Substitute $x+5$ for $y$ in equation 1.
      $\begin{array}{lllllllllll}\hfill x+\left(x+5\right)& =\hfill & 37\hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill x+x+5& =\hfill & 37\hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill 2x+5& =\hfill & 37\hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill 2x& =\hfill & 32\hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill x& =\hfill & 16\hfill & \hfill & \text{Then,}\hfill & \hfill & y\hfill & =\hfill & 16+5\hfill & =\hfill & 21.\hfill \\ \hfill x& =\hfill & 16,\hfill & \hfill & \hfill & \hfill & y\hfill & =\hfill & 21\hfill & \hfill & \hfill \end{array}$

Step 4:  The Sum is 37.
      $\begin{array}{rrrr}\hfill x+y& \hfill =& \hfill 37& \hfill \\ \hfill 16+21& \hfill =& \hfill 37& \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ \hfill 37& \hfill =& \hfill 37& \text{Yes,\hspace{0.17em}this\hspace{0.17em}is\hspace{0.17em}correct}\text{.}\hfill \end{array}$
 One is 5 larger than the other.
      $\begin{array}{rrrr}\hfill y& \hfill =& x+5\hfill & \hfill \\ \hfill 21& \hfill =& 16+5\hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ \hfill 21& \hfill =& 21\hfill & \text{Yes,\hspace{0.17em}this\hspace{0.17em}is\hspace{0.17em}correct}\text{.}\hfill \end{array}$

Step 5:  The two numbers are 16 and 21.

A parking meter contains 27 coins consisting only of dimes and quarters. If the meter contains $4.35, how many of each type of coin is there?

$\begin{array}{llll}\text{Step 1:}\hfill & \text{Let\hspace{0.17em}}D\hfill & =\hfill & \text{number\hspace{0.17em}of\hspace{0.17em}dimes}.\hfill \\ \hfill & Q\hfill & =\hfill & \text{number\hspace{0.17em}of\hspace{0.17em}quarters}.\hfill \end{array}$

Step 2:  There are two relationships.
     (a)  There are 27 coins.   $D+Q=27.$
     (b) Contribution due to dimes =
       Contribution due to quarters =

        $\begin{array}{l}10D+25Q=435\\ \begin{array}{ccc}\{\begin{array}{c}D+Q=27\\ 10D+25Q=435\end{array}& & \begin{array}{l}\left(1\right)\\ \left(2\right)\end{array}\end{array}\end{array}$

Step 3:   $\begin{array}{lll}\{\begin{array}{l}D+Q=27\\ 10D+25Q=435\end{array}\hfill & \hfill & \begin{array}{l}\left(1\right)\\ \left(2\right)\end{array}\hfill \end{array}$
 We can solve this system using elimination by addition. Multiply both sides of equation $\left(1\right)$ by $-10$ and add.

      $\begin{array}{l}\begin{array}{ccc}\frac{\begin{array}{l}-10D-10Q=-270\\ 10D+25Q=435\end{array}}{\begin{array}{l}15Q=165\\ Q=11\end{array}}& & \begin{array}{l}\\ \\ \\ \text{Then,\hspace{0.17em}}D+11=27\\ D=16\end{array}\end{array}\\ D=16,Q=11\end{array}$

Step 4:  16 dimes and 11 quarters is 27 coins.
      $\begin{array}{rrrr}\hfill 10\left(16\right)+11\left(25\right)& \hfill =& \hfill 435& \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ \hfill 160+275& \hfill =& \hfill 435& \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ \hfill 435& \hfill =& \hfill 435& \text{Yes,\hspace{0.17em}this\hspace{0.17em}is\hspace{0.17em}correct}\text{.}\hfill \end{array}$
 The solution checks.

Step 5:  There are 11 quarters and 16 dimes.