# Solve the equations and inequalities. Write the solution sets to the inequalities in interval notation. 5<=3+|2x-7|

Question
Solve the equations and inequalities. Write the solution sets to the inequalities in interval notation.
$$\displaystyle{5}\le{3}+{\left|{2}{x}-{7}\right|}$$

2020-12-17
Simplify the give expression as follows.
$$\displaystyle{3}+{\left|{2}{x}-{7}\right|}\ge{5}$$
$$\displaystyle{3}+{\left|{2}{x}-{7}\right|}-{3}\ge{5}-{3}$$ (subtract 3 from both sides)
$$\displaystyle{\left|{2}{x}-{7}\right|}\ge{2}$$
by the absolute rule, if $$\displaystyle{\left|{u}\right|}\ge{a}$$, a>0 then $$\displaystyle{\left|{u}\right|}\le-{a}{\quad\text{and}\quad}{\left|{u}\right|}\ge{a}$$
$$\displaystyle{2}{x}-{7}\le-{2}{\quad\text{and}\quad}{2}{x}-{7}\ge{2}$$ ($$\displaystyle\because$$ by absolute rule)
$$\displaystyle{2}{x}-{7}+{7}\le-{2}{\quad\text{and}\quad}{2}{x}-{7}+{7}\ge{2}+{7}$$ (add 7 on both sides)
$$\displaystyle\frac{{{2}{x}}}{{2}}\le\frac{{5}}{{2}}{\quad\text{and}\quad}\frac{{{2}{x}}}{{2}}\ge\frac{{9}}{{2}}$$ (divie by 2 onn both sides)
$$\displaystyle{x}\le\frac{{5}}{{2}}{\quad\text{and}\quad}{x}\ge\frac{{9}}{{2}}$$
Therefore, the solution set of the given inequaliry is $$\displaystyle{\left(-\infty,\frac{{5}}{{2}}\right]}\cup{\left[\frac{{9}}{{2}},\infty\right)}$$.

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