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Review Quiz tomorrow!
Homework: bring in all review homework (review 3-6) completed tomorrow. Also, write a justification for every step on this sheet.
Simplifying expressions: If $x = - 3$ , find $\frac{- 3 x^{2} - x^{3} + | x |}{x^{2} - 6}$ .

Solution: $\begin{array}{lll} \frac{- 3 x^{2} - x^{3} + | x |}{x^{2} - 6} & = & \frac{- 3 {(- 3)}^{2} - {(- 3)}^{3} + | - 3 |}{{(- 3)}^{2} - 6} \\ = & \frac{- 3 (9) - (- 27) + 3}{9 - 6} \\ = & \frac{- 27 + 27 + 3}{3} \\ = & \frac{3}{3} \\ = & 1 \end{array}$
Simplify the following expression into a single fraction.
$4 - \frac{2}{3} (\frac{x^{2}}{2} - \frac{3}{5})$ $\begin{array}{rcl} = & 4 - \frac{2}{3} (\frac{x^{2}}{2}) - (- \frac{2}{3}) (\frac{3}{5}) \\ = & 4 - \frac{2 x^{2}}{6} - (- \frac{6}{15}) \\ = & 4 - \frac{2 x^{2}}{6} + \frac{6}{15} \\ = & \frac{4 \cdot 30}{30} - \frac{5 \cdot 2 x^{2}}{5 \cdot 6} + \frac{2 \cdot 6}{2 \cdot 15} \\ = & \frac{120}{30} - \frac{10 x^{2}}{30} + \frac{12}{30} \\ = & \frac{132 - 10 x^{2}}{30} \end{array}$
Solve: $- \frac{3}{5} x - \frac{2}{3} = 2 + \frac{5}{3} x$
$\begin{array}{rcl} - \frac{3}{5} x - \frac{2}{3} & = & 2 + \frac{5}{3} x \\ - \frac{3}{5} x - \frac{5}{3} x & = & 2 + \frac{2}{3} \\ - \frac{3 \cdot 3}{3 \cdot 5} x - \frac{5 \cdot 5}{5 \cdot 3} x & = & \frac{3 \cdot 2}{3} + \frac{2}{3} \\ - \frac{9}{15} x - \frac{25}{15} x & = & \frac{6}{3} + \frac{2}{3} \\ - \frac{34}{15} x & = & \frac{8}{3} \\ - \frac{15}{34} \cdot (- \frac{34}{15}) x & = & - \frac{15}{34} \cdot \frac{8}{3} \\ x & = & - \frac{120}{102} \\ = & - \frac{20}{17} \end{array}$
How many mm in a yard? (Note: .0254 m = 1 in)

Solution: $\frac{1000 mm}{1 m} \cdot \frac{. 0254 m}{1 in} \cdot \frac{36 in}{yd} = \frac{914.4 mm}{yd}$
Solve the inequality and graph the solution (or write “no solution”):
$4 x + 5 > 3 (x - 2) and - 2 x \geq - 6$ $4 x + 5 > 3 x - 6 and x \leq 3$ $x > - 11 and x \leq 3$ $- 11 < x and x \leq 3$ $- 11 < x \leq 3$
You are running a race. After $6 \frac{1}{2}$ minutes you are four miles from the finish line. After $19 \frac{1}{2}$ minutes you are two miles from the finish line. Assuming that you run at a constant speed, what is your speed? Write an equation that relates how far you are from the finish line, $y$ , with the number of minutes after the race starts, $x$ . How far were you when the race started? What time will you finish?

Solution:
- Speed is rate of change or slope of the graph: $\frac{2 - 4}{19 \frac{1}{2} - 6 \frac{1}{2}} = \frac{- 2}{13}$
- Using point/slope form, $y - 4 = \frac{- 2}{13} (x - 6 \frac{1}{2})$
- How far were we when the race started? Then $x = 0$ so we solve $y - 4 = \frac{- 2}{13} (0 - 6 \frac{1}{2})$ :
  $\begin{array}{rcl} y - 4 & = & \frac{- 2}{13} (0 - 6 \frac{1}{2}) \\ = & \frac{13}{13} \\ = & 1 \\ y & = & 1 + 4 \\ = & 5 \end{array}$
- What time will you finish? In this case $y = 0$ so
  $\begin{array}{rcl} 0 - 4 & = & \frac{- 2}{13} (x - 6 \frac{1}{2}) \\ - 4 & = & - \frac{2}{13} x + \frac{13}{13} \\ = & - \frac{2}{13} x + 1 \\ - 5 & = & - \frac{2}{13} x \\ (- \frac{13}{2}) (- 5) & = & (- \frac{13}{2}) (- \frac{2}{13} x) \\ \frac{65}{2} & = & x \\ 32 \frac{1}{2} & = & x \end{array}$