How do you solve 3x-8>=1/2 (10x+7)?

Jul 29, 2017

See a solution process below:

Explanation:

First, expand the term on the right side of the inequality by multiplying each term within the parenthesis by the term outside the parenthesis:

$3 x - 8 \ge \textcolor{red}{\frac{1}{2}} \left(10 x + 7\right)$

$3 x - 8 \ge \left(\textcolor{red}{\frac{1}{2}} \times 10 x\right) + \left(\textcolor{red}{\frac{1}{2}} \times 7\right)$

$3 x - 8 \ge \frac{10}{2} x + \frac{7}{2}$

$3 x - 8 \ge 5 x + \frac{7}{2}$

Next, subtract $\textcolor{red}{3 x}$ and $\textcolor{b l u e}{\frac{7}{2}}$ from each side of the inequality to isolate the $x$ term while keeping the inequality balanced:

-color(red)(3x) + 3x - 8 - color(blue)(7/2) >= -color(red)(3x) + 5x + 7/2 - color(blue)(7/2)

$0 - \left(\frac{2}{2} \times 8\right) - \textcolor{b l u e}{\frac{7}{2}} \ge \left(- \textcolor{red}{3} + 5\right) x + 0$

$- \frac{16}{2} - \textcolor{b l u e}{\frac{7}{2}} \ge 2 x$

$- \frac{23}{2} \ge 2 x$

Now. multiply each side of the inequality by $\textcolor{red}{\frac{1}{2}}$ to solve for $x$ while keeping the inequality balanced:

$\textcolor{red}{\frac{1}{2}} + - \frac{23}{2} \ge \textcolor{red}{\frac{1}{2}} \times 2 x$

$- \frac{23}{4} \ge \textcolor{red}{\frac{1}{\textcolor{b l a c k}{\cancel{\textcolor{red}{2}}}}} \times \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} x$

$- \frac{23}{4} \ge x$

To state the solution in terms of $x$ we can reverse or "flip" the entire inequality:

$x \le - \frac{23}{4}$