5.5 Application ii - solving problems

This year an item costs $\$44$ , an increase of $\$3$ over last year’s price. What was last year’s price?

$\begin{array}{l}\begin{array}{lllll}\text{Step}1:\hfill & \hfill \text{Let}x& =\hfill & \text{last}\text{year's}\text{price}\text{.}\hfill & \hfill \\ \text{Step}2:\hfill & \hfill x+3& =\hfill & 44.\hfill & x+3\text{represents}\text{the}\text{\$3}\text{increase}\text{in}\text{price}\text{.}\hfill \\ \text{Step}3:\hfill & \hfill x+3& =\hfill & 44\hfill & \hfill \\ \hfill & \hfill x+3-3& =\hfill & 44-3\hfill & \hfill \\ \hfill & \hfill x& =\hfill & 41\hfill & \hfill \\ \text{Step}4:\hfill & \hfill 41+3& =\hfill & \text{44}\hfill & \text{Yes,}\text{this}\text{is}\text{correct}\text{.}\hfill \end{array}\\ \begin{array}{ll}\text{Step}5:\hfill & \text{Last}\text{year's}\text{price}\text{was}\$41.\hfill \end{array}\end{array}$

The perimeter (length around) of a square is 60 cm (centimeters). Find the length of a side.

$\begin{array}{ll}\text{Step}\text{1:}\hfill & \text{Let}x\text{=}\text{length}\text{of}\text{a}\text{side}\text{.}\hfill \\ \text{Step}\text{2:}\hfill & \text{We}\text{can}\text{draw}\text{a}\text{picture}\text{.}\hfill \end{array}$

$\begin{array}{l}\begin{array}{lllll}\text{Step}3:\hfill & \hfill x+x+x+x& =\hfill & 60\hfill & \hfill \\ \hfill & \hfill 4x& =\hfill & 60\hfill & \text{Divide}\text{both}\text{sides}\text{by}4.\hfill \\ \hfill & \hfill x& =\hfill & 15.\hfill & \hfill \\ \text{Step}4:\hfill & \hfill 4(15)& =\hfill & 60.\hfill & \hfill \text{Yes,}\text{this}\text{is}\text{correct}\text{.}\end{array}\\ \begin{array}{ll}\text{Step}5:\hfill & \text{The}\text{length}\text{of}\text{a}\text{side}\text{is}15\text{cm}\text{.}\hfill \end{array}\end{array}$

Six percent of a number is 54. What is the number?

$\begin{array}{ll}\text{Step}1:\hfill & \text{Let}x=\text{the}\text{number}\hfill \\ \text{Step}2:\hfill & \text{We}\text{must}\text{convert}6\%\text{to}\text{a}\text{decimal.}\hfill \end{array}$
$\begin{array}{l}\begin{array}{lllll}\hfill & \hfill 6\%& =\hfill & .06\hfill & \hfill \\ \hfill & \hfill .06x& =\hfill & 54\hfill & .06x\text{occurs}\text{because}\text{we}\text{want}6\%\text{of}x\text{.}\hfill \\ \text{Step}3:\hfill & \hfill .06x& =\hfill & 54.\hfill & \text{Divide}\text{both}\text{sides}\text{by}\mathrm{.06.}\hfill \\ \hfill & \hfill x& =\hfill & \frac{54}{.06}\hfill & \hfill \\ \hfill & \hfill x& =\hfill & 900\hfill & \hfill \\ \text{Step}4:\hfill & .06(900)\hfill & =\hfill & 54.\hfill & \text{Yes,}\text{this}\text{is}\text{correct}\text{.}\hfill \end{array}\\ \begin{array}{ll}\text{Step}5:\hfill & \text{The}\text{number}\text{is}900.\hfill \end{array}\end{array}$

An astronomer notices that one star gives off about $3.6$ times as much energy as another star. Together the stars give off $55.844$ units of energy. How many units of energy does each star emit?

1. In this problem we have two unknowns and, therefore, we might think, two variables. However, notice that the energy given off by one star is given in terms of the other star. So, rather than introducing two variables, we introduce only one. The other unknown(s) is expressed in terms of this one. (We might call this quantity the base quantity.)

   Let $x=$ number of units of energy given off by the less energetic star. Then, $3.6x=$ number of units of energy given off by the more energetic star.

   $\begin{array}{l}\begin{array}{lllll}\text{Step}\text{2:}\hfill & \hfill x+3.6x& \text{=}\hfill & \mathrm{55.844.}\hfill & \hfill \\ \text{Step}\text{3:}\hfill & \hfill x+3.6x& \text{=}\hfill & 55.844\hfill & \hfill \\ \hfill & \hfill 4.6x& \text{=}\hfill & 55.844\hfill & \text{Divide}\text{both}\text{sides}\text{by}\text{4}\text{.6}\text{.}\text{A}\text{calculator}\text{would}\text{be}\text{useful}\text{at}\text{this}\hfill \\ \hfill & \hfill & \hfill & \hfill & \text{point}\text{.}\hfill \\ \hfill & \hfill x& \text{=}\hfill & \frac{55.844}{4.6}\hfill & \hfill \\ \hfill & \hfill x& \text{=}\hfill & 12.14\hfill & \text{The}\text{wording}\text{of}\text{the}\text{problem}\text{implies}two\text{numbers}\text{are}\text{needed}\hfill \\ \hfill & \hfill & \text{=}\hfill & \hfill & \text{for}\text{a}\text{complete}\text{solution}\text{.}\text{We}\text{need}\text{the}\text{number}\text{of}\text{units}\text{of}\hfill \\ \hfill & \hfill & \hfill & \hfill & \text{energy}\text{for}\text{the}\text{other}\text{star.}\hfill \\ \hfill & \hfill 3.6x& \text{=}\hfill & 3.6(12.14)\hfill & \hfill \\ \hfill & \hfill & \text{=}\hfill & 43.704\hfill & \hfill \\ \text{Step}\text{4:}\hfill & \hfill 12.14+43.704& \text{=}\hfill & \mathrm{55.844.}\hfill & \text{Yes,}\text{this}\text{is}\text{correct}\text{.}\hfill \end{array}\\ \begin{array}{ll}\text{Step}5:\hfill & \text{One}\text{star}\text{gives}\text{off}12.14\text{units}\text{}\text{of}\text{energy}\text{and}\text{the}\text{other}\text{star}\text{gives}\text{off}43.704\text{units}\text{of}\text{energy}\text{.}\hfill \end{array}\end{array}$