Absolute Value (in equations)
Find the value(s) of $x$:
- $|x+1|=2 \rightarrow x+1=2 \text{ and } x+1=-2\rightarrow \mathbf{x=1,-3}$
- $|x-5|=-5 \rightarrow \mathbf{\text{no solution}}$
- $|x-5|=5 \rightarrow x-5=5 \text{ and } x-5=-5\rightarrow\mathbf{x=10,0}$
- $|5x|=10 \rightarrow 5x=10 \text{ and } 5x=-10\rightarrow\mathbf{x=2,-2}$
- $|2x|=3 \rightarrow 2x=3 \text{ and }2x=-3\rightarrow\mathbf{x=\dfrac{3}{2},\dfrac{-3}{2}}$
- $|\dfrac{x}{2}|=6 \rightarrow \dfrac{x}{2}=6 \text{ and } \dfrac{x}{2}=-6\rightarrow\mathbf{x=12,-12}$
- $|\dfrac{x}{3}|=10\rightarrow \dfrac{x}{3}=10 \text{ and } \dfrac{x}{3}=-10\rightarrow\mathbf{x=30,-30}$
- $|\dfrac{1}{x}|=2 \rightarrow \dfrac{1}{x}=2 \text{ and } \dfrac{1}{x}=-2\rightarrow\mathbf{x=\dfrac{1}{2},\dfrac{-1}{2}}$
- $|3x+4|=12\rightarrow 3x+4=12 \text{ and }3x+4=-12\rightarrow\mathbf{x=\dfrac{8}{3},\dfrac{-16}{3}}$
- $|12x-7|=20\rightarrow 12x-7=20 \text{ and }12x-7=-20\rightarrow\mathbf{x=\dfrac{9}{4},\dfrac{-13}{12}}$
- $|\dfrac{2x+1}{2}|=10 \rightarrow \dfrac{2x+1}{2}=10 \text{ and } \dfrac{2x+1}{2}=-10\rightarrow\mathbf{x=\dfrac{19}{2},\dfrac{-21}{2}}$
- $|\dfrac{9x-10}{5}|=3 \rightarrow \dfrac{9x-10}{5}=3 \text{ and } \dfrac{9x-10}{5}=-3\rightarrow\mathbf{x=\dfrac{25}{9},\dfrac{-5}{9}}$