Practice Solutions

1. $2x{y}^2+3x^{3}y$

$$\begin{gathered} 2\left(–1\right){\left(–2\right)}^2+3{\left(–1\right)}^3\left(–2\right) \\ 2\left(–1\right)\left(4\right)+3\left(–1\right)\left(–2\right) \\ –8+6 \\ \overline{)–2} \end{gathered}$$

2. $–3y+7x–xy$

$$\begin{gathered} –3\left(–2\right)+7\left(–1\right)–\left(–1\right)\left(–2\right) \\ 6–7–2 \\ –1–2 \\ \overline{)–3} \end{gathered}$$

3. $–ab+\frac{b}{a}$

$$\begin{gathered} –\left(5\right)\left(15\right)+\frac{15}{5} \\ –75+3 \\ \overline{)–72} \end{gathered}$$

4. $–\frac{1}{3}b+a$

$$\begin{gathered} –\frac{1}{3}\left(15\right)+5 \\ –5+5 \\ \overline{)0} \end{gathered}$$

5. $\frac{1}{4}{x}^2 {\left(xy\right)}^2$

$$\begin{gathered} \frac{1}{4}{\left(2\right)}^2 {\left(\left(2\right)\left(–3\right)\right)}^2 \\ \frac{1}{4}\left(4\right){\left(–6\right)}^2 \\ 1\left(36\right) \\ \overline{)36} \end{gathered}$$

6. $\frac{4}{xy}÷\left(x^{2}y\right)$

$$\begin{gathered} \frac{4}{\left(2\right)\left(–3\right)}÷\left(2^2·–3\right) \\ \frac{–4}{6}÷\left(4·–3\right)=\frac{–2}{3}÷–12 \\ \frac{{}^1\cancel{–2}}{3}·\frac{1}{{\cancel{–12}}_6}=\overline{)\frac{1}{18}} \end{gathered}$$

7. $ab+bc$

$$\begin{gathered} 4\left(5\right)+5\left(–1\right) \\ 20–5 \\ \overline{)15} \end{gathered}$$

8. $\frac{b}{a}–\frac{c}{a}$

$$\begin{gathered} \frac{5}{4}–\frac{–1}{4} \\ \frac{5}{4}+\frac{1}{4}=\frac{6}{4}=\overline{)\frac{3}{2}} \end{gathered}$$

9. $–3x+16=43 \mathrm{when} x=9 \mathrm{or} x=–9$

$\begin{array}{rcl}& & \begin{array}{ccc}–3\left(9\right)+16=43& & –3\left(–9\right)+16=43\\ –27+16=43& & 27+16=43\\ –11=43 \mathbf{☒}& & 43=43 ✓\end{array}\\ x& =& –9\end{array}$

10. $\frac{x}{4}–7=–11 \mathrm{when} x=–16 \mathrm{or} x=1$

$\begin{array}{ccc}\frac{\left(–16\right)}{4}–7=–11& & \frac{\left(1\right)}{4}–7=–11\\ –4–7=–11& & \frac{1}{4}–\frac{28}{4}=–11\\ –11=–11 ✓& & –\frac{27}{4}=–11 \begin{array}{rcl}& & \mathbf{☒}\end{array}\\ & & \end{array}$

11. $\frac{5}{2}\left(3x+2\right)=9 \mathrm{when} x=\frac{8}{15} \mathrm{or} x=30$

$\begin{array}{ccc}\frac{5}{2}\left(3\left(\frac{8}{15}\right)+2\right)=9& & \frac{5}{2}\left(3\left(30\right)+2\right)=9\\ \frac{5}{2}\left(\frac{8}{5}+2\right)=9& & \frac{5}{2}\left(90+2\right)=9\\ \frac{40}{10}+\frac{10}{2}=9& & \frac{5}{2}\left(92\right)=9 \\ 4+5=9 & & 230=9 \begin{array}{rcl}& & \mathbf{☒}\end{array}\\ 9=9 ✓& & \end{array}$

12. $12–7x=–2 \mathrm{when} x=–2 \mathrm{or} x=2$

$$\begin{gathered} \begin{array}{ccc}12–7\left(–2\right)=–2& & 12–7\left(2\right)=–2\\ 12+14=–2& & 12–14=–2\\ 28=–2\begin{array}{rcl}& & \mathbf{ }\mathbf{☒}\end{array} & & –2=–2 ✓\end{array} \\ x=2 \end{gathered}$$