How do you solve and write the following in interval notation: 5(x − 3) ≤ −2(x + 1)?

Mar 11, 2018

See a solution process below:

Explanation:

To solve this, first, expand the terms in parenthesis by multiplying each term with the parenthesis by the term outside the parenthesis being careful to manage the signs correctly:

$\textcolor{red}{5} \left(x - 3\right) \le \textcolor{b l u e}{- 2} \left(x + 1\right)$

$\left(\textcolor{red}{5} \times x\right) - \left(\textcolor{red}{5} \times 3\right) \le \left(\textcolor{b l u e}{- 2} \times x\right) + \left(\textcolor{b l u e}{- 2} \times 1\right)$

$5 x - 15 \le - 2 x + \left(- 2\right)$

$5 x - 15 \le - 2 x - 2$

Next, add $\textcolor{red}{15}$ and $\textcolor{b l u e}{2 x}$ to each side of the inequality to isolate the $x$ term while keeping the inequality balanced:

$5 x + \textcolor{b l u e}{2 x} - 15 + \textcolor{red}{15} \le - 2 x + \textcolor{b l u e}{2 x} - 2 + \textcolor{red}{15}$

$\left(5 + \textcolor{b l u e}{2}\right) x - 0 \le 0 + 13$

$7 x \le 13$

Now, divide each side of the inequality by $\textcolor{red}{7}$ to solve for $x$ while keeping the inequality balanced:

$\frac{7 x}{\textcolor{red}{7}} \le \frac{13}{\textcolor{red}{7}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}} x}{\cancel{\textcolor{red}{7}}} \le \frac{13}{7}$

$x \le \frac{13}{7}$

We can write this in interval notation as:

$\left(- \infty , \frac{13}{7}\right]$