###### Example3.6.12

Solve for $$z$$ in the inequality $$\frac{3z}{5}+\frac{1}{2}\le\left(\frac{z}{10}+\frac{3}{4}\right)+\left(\frac{z}{2}-\frac{1}{4}\right)\text{.}$$

To solve for $$z\text{,}$$ we will first need to multiply each side of the inequality by the LCD, which is $$40\text{.}$$ After that, we'll finish solving by putting all terms containing a variable on one side of the inequality:

\begin{align*} \frac{3z}{5}+\frac{1}{2}\amp\le\left(\frac{z}{10}+\frac{3}{4}\right)+\left(\frac{z}{2}-\frac{1}{4}\right)\\ \multiplyleft{40}\left(\frac{3z}{5}+\frac{1}{2}\right)\amp\le \multiplyleft{40}\left(\left(\frac{z}{10}+\frac{3}{4}\right)+\left(\frac{z}{2}-\frac{1}{4}\right)\right)\\ \multiplyleft{40}\left(\frac{3z}{5}\right)+\multiplyleft{40}\left(\frac{1}{2}\right)\amp\le \multiplyleft{40}\left(\frac{z}{10}+\frac{3}{4}\right)+\multiplyleft{40}\left(\frac{z}{2}-\frac{1}{4}\right)\\ 40\cdot\left(\frac{3z}{5}\right)+40\cdot\left(\frac{1}{2}\right)\amp\le \multiplyleft{40}\left(\frac{z}{10}\right)+\multiplyleft{40}\left(\frac{3}{4}\right)+\multiplyleft{40}\left(\frac{z}{2}\right)-\multiplyleft{40}\left(\frac{1}{4}\right)\\ 24z+20\amp\le 4z+30+20z-10\\ 24z+20\amp\le 24z+20\\ 24z+20\subtractright{24z}\amp\le 24z+20\subtractright{24z}\\ 20\amp\le20 \end{align*}

As the equation $$20\le20$$ is true for all values of $$z\text{,}$$ all real numbers are solutions to this inequality. Thus the solution set is $$\{z\mid z\text{ is a real number}\}\text{.}$$

in-context