Question

Solve.

(i) $\left|3x-5\right|=1$ (ii) $\left|7-2x\right|=5$ (iii) $\left|\frac{8-x}{2}\right|=5$ (iv) $\left|5+\frac{x}{4}\right|=5$

Answer 1

(i) $\left|3x-5\right|=1$
$$\begin{gathered} \Rightarrow 3x-5=\pm 1 \\ \Rightarrow 3x-5=1,\mathrm{or},3x-5=-1 \\ \Rightarrow 3x=1+5,\mathrm{or},3x=-1+5 \\ \Rightarrow 3x=6,\mathrm{or},3x=4 \\ \therefore x=2,\mathrm{or},x=\frac{4}{3} \end{gathered}$$

(ii) $\left|7-2x\right|=5$
$$\begin{gathered} \Rightarrow 7-2x=\pm 5 \\ \Rightarrow 7-2x=5,\mathrm{or},7-2x=-5 \\ \Rightarrow 2x=7-5,\mathrm{or},2x=7+5 \\ \Rightarrow 2x=2,\mathrm{or},2x=12 \\ \therefore x=1,\mathrm{or},x=6 \end{gathered}$$

(iii) $\left|\frac{8-x}{2}\right|=5$
$$\begin{gathered} \Rightarrow \frac{8-x}{2}=\pm 5 \\ \Rightarrow \frac{8-x}{2}=5,\mathrm{or},\frac{8-x}{2}=-5 \\ \Rightarrow 8-x=10,\mathrm{or},8-x=-10 \\ \Rightarrow x=8-10,\mathrm{or},x=8+10 \\ \therefore x=-2,\mathrm{or},x=18 \end{gathered}$$

(iv) $\left|5+\frac{x}{4}\right|=5$
$$\begin{gathered} \Rightarrow 5+\frac{x}{4}=\pm 5 \\ \Rightarrow 5+\frac{x}{4}=5,\mathrm{or},5+\frac{x}{4}=-5 \\ \Rightarrow \frac{x}{4}=5-5,\mathrm{or},\frac{x}{4}=-5-5 \\ \Rightarrow \frac{x}{4}=0,\mathrm{or},\frac{x}{4}=-10 \\ \therefore x=0,\mathrm{or},x=-40 \end{gathered}$$