Review Lesson 1: Practice Solutions

Solve.

Note

Not showing that you have checked your work does not make your solution incorrect. However, it is critical to make a habit of checking your work, either on paper or using technology.

$\frac{5}{8} x - \frac{8}{7} = \frac{15}{7}$

$\begin{array}{rcl} LCD (8, 7) & = & 56 \\ (56) (\frac{5}{8} x - \frac{8}{7}) & = & (\frac{15}{7}) (56) \\ 35 x - 64 & = & 120 \\ + 64 + 64 \\ (\frac{1}{35}) 35 x & = & 184 (\frac{1}{35}) \\ x & = & \frac{184}{35} \end{array}$

Check

$\frac{5}{8} (\frac{184}{35}) - \frac{8}{7} = \frac{15}{7} \frac{23}{7} - \frac{8}{7} = \frac{15}{7} ✓$

$5 = \frac{3 (4 - x)}{4}$

$\begin{array}{rcl} (4) 5 & = & \frac{3 (4 - x)}{4} (4) \\ 20 & = & 3 (4 - x) \\ 20 & = & 12 - 3 x \\ - 12 - 12 \\ (- \frac{1}{3}) 8 & = & - 3 x (- \frac{1}{3}) \\ x & = & - \frac{8}{3} \end{array}$

Check

$5 = \frac{3 (4 - (- \frac{8}{3}))}{4} 5 = \frac{12 + 8}{4} 5 = \frac{20}{4} ✓$

$\frac{5}{2} + \frac{x}{6} = 4 x$

$\begin{array}{rcl} LCD (2, 6) & = & 6 \\ (6) (\frac{5}{2} + \frac{x}{6}) & = & 4 x (6) \\ 15 + x & = & 24 x \\ - x - x \\ (\frac{1}{23}) 15 & = & 23 x (\frac{1}{23}) \\ x & = & \frac{15}{23} \end{array}$

Check

$\frac{5}{2} + \frac{(\frac{15}{23})}{6} = 4 (\frac{15}{23}) \frac{5}{2} + \frac{5}{46} = \frac{60}{23} \frac{120}{46} + \frac{5}{46} = \frac{60}{23} ✓$

$- 6 (\frac{2}{3} x - 1) = \frac{1}{2} (7 - 3 x)$

$\begin{array}{rcl} - 4 x + 6 & = & \frac{7}{2} - \frac{3}{2} x \\ LCD & = & 2 \\ (2) (- 4 x + 6) & = & (\frac{7}{2} - \frac{3}{2} x) (2) \\ - 8 x + 12 & = & 7 - 3 x \\ + 8 x - 7 - 7 + 8 x \\ (\frac{1}{5}) 5 & = & 5 x (\frac{1}{5}) \\ x & = & 1 \end{array}$

Check

$- 6 (\frac{2}{3} (1) - 1) = \frac{1}{2} (7 - 3 (1)) - 6 (\frac{2}{3} - 1) = \frac{1}{2} (4) - 6 (- \frac{1}{3}) = 2 ✓$

Solve.

$\frac{x + 14}{8} = 2 (x - 1)$

$\begin{array}{rcl} (8) \frac{x + 14}{8} & = & (2 x - 2) (8) \\ x + 14 & = & 16 x - 16 \\ - x + 16 - x + 16 \\ (\frac{1}{15}) 30 & = & 15 x (\frac{1}{15}) \\ x & = & 2 \end{array}$

Check

$\frac{(2) + 14}{8} = 2 ((2) - 1) \frac{16}{8} = 2 (1) ✓$

$\frac{7}{8} x - \frac{2}{3} = \frac{5}{6} x - \frac{3}{4}$

$\begin{array}{rcl} LCD (8, 3, 6, 4) & = & 24 \\ (24) (\frac{7}{8} x - \frac{2}{3}) & = & (\frac{5}{6} x - \frac{3}{4}) (24) \\ 21 x - 16 & = & 20 x - 18 \\ - 20 x + 16 - 20 x + 16 \\ x & = & - 2 \end{array}$

Check

$\frac{7}{8} (- 2) - \frac{2}{3} = \frac{5}{6} (- 2) - \frac{3}{4} - \frac{14}{8} - \frac{2}{3} = - \frac{10}{6} - \frac{3}{4} - \frac{42}{24} - \frac{16}{24} = - \frac{40}{24} - \frac{18}{24} ✓$

$\frac{3}{7} (1 + x) - \frac{x}{7} + 1 = 1$

$\begin{array}{rcl} LCD & = & 7 \\ (7) (\frac{3}{7} (1 + x) - \frac{x}{7} + 1) & = & 1 (7) \\ 3 (1 + x) - x + 7 & = & 7 \\ 3 + 3 x - x + 7 & = & 7 \\ 2 x + 10 & = & 7 \\ - 10 - 10 \\ (\frac{1}{2}) 2 x & = & - 3 (\frac{1}{2}) \\ x & = & - \frac{3}{2} \end{array}$

Check

$\frac{3}{7} (1 + (- \frac{3}{2})) - \frac{(- \frac{3}{2})}{7} + 1 = 1 \frac{3}{7} (- \frac{1}{2}) + \frac{3}{14} + 1 = 1 - \frac{3}{14} + \frac{3}{14} + 1 = 1 ✓$

$9 x - 5 = - \frac{3}{4} (3 - x)$

$\begin{array}{rcl} LCD & = & 4 \\ (4) (9 x - 5) & = & (- \frac{3}{4} (3 - x)) (4) \\ 36 x - 20 & = & - 3 (3 - x) \\ 36 x - 20 & = & - 9 + 3 x \\ - 3 x + 20 + 20 - 3 x \\ (\frac{1}{33}) 33 x & = & 11 (\frac{1}{33}) \\ x & = & \frac{11}{33} = \frac{1}{3} \end{array}$

Check

$9 (\frac{1}{3}) - 5 = - \frac{3}{4} (3 - (\frac{1}{3})) 3 - 5 = - \frac{3}{4} (\frac{9}{3} - \frac{1}{3}) - 2 = - \frac{3}{4} (\frac{8}{3}) ✓$