# How do you solve and write the following in interval notation: -5(13x + 3) < - 2(13x - 3)?

Jul 8, 2017

See a solution process below:

#### Explanation:

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

$\textcolor{red}{- 5} \left(13 x + 3\right) < \textcolor{b l u e}{- 2} \left(13 x - 3\right)$

$\left(\textcolor{red}{- 5} \times 13 x\right) + \left(\textcolor{red}{- 5} \times 3\right) < \left(\textcolor{b l u e}{- 2} \times 13 x\right) - \left(\textcolor{b l u e}{- 2} \times 3\right)$

$- 65 x + \left(- 15\right) < - 26 x - \left(- 6\right)$

$- 65 x - 15 < - 26 x + 6$

Next, add $\textcolor{red}{65 x}$ and subtract $\textcolor{b l u e}{6}$ from each side of the inequality to isolate the $x$ term while keeping the inequality balanced:

color(red)(65x) - 65x - 15 - color(blue)(6) < color(red)(65x) - 26x + 6 - color(blue)(6)

$0 - 21 < \left(\textcolor{red}{65} - 26\right) x + 0$

$- 21 < 39 x$

Now, divide each side of the inequality by $\textcolor{red}{39}$ to solve for $x$ while keeping the inequality balanced:

$- \frac{21}{\textcolor{red}{39}} < \frac{39 x}{\textcolor{red}{39}}$

$- \frac{3 \times 7}{\textcolor{red}{3 \times 13}} < \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{39}}} x}{\cancel{\textcolor{red}{39}}}$

$- \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} \times 7}{\textcolor{red}{\textcolor{b l a c k}{\cancel{\textcolor{red}{3}}} \times 13}} < x$

$- \frac{7}{13} < x$

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

$x > - \frac{7}{13}$