# How do you solve the inequality 3(t - 3) + 1 ≥ 7 and 2(t + 1) + 3 ≤ 1?

Aug 6, 2017

See a solution process below:

#### Explanation:

Solve First Inequality

Begin by solving the first inequality:
$\textcolor{red}{3} \left(t - 3\right) + 1 \ge 7$

$\left(\textcolor{red}{3} \times t\right) - \left(\textcolor{red}{3} \times 3\right) + 1 \ge 7$

$3 t - 9 + 1 \ge 7$

$3 t - 8 \ge 7$

$3 t - 8 + \textcolor{red}{8} \ge 7 + \textcolor{red}{8}$

$3 t - 0 \ge 15$

$3 t \ge 15$

$\frac{3 t}{\textcolor{red}{3}} \ge \frac{15}{\textcolor{red}{3}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} t}{\cancel{\textcolor{red}{3}}} \ge 5$

$t \ge 5$

Solve Second Inequality

Next, we can solve the second inequality for $t$:

$\textcolor{red}{2} \left(t + 1\right) + 3 \le 1$

$\left(\textcolor{red}{2} \times t\right) + \left(\textcolor{red}{2} \times 1\right) + 3 \le 1$

$2 t + 2 + 3 \le 1$

$2 t + 5 \le 1$

$2 t + 5 - \textcolor{red}{5} \le 1 - \textcolor{red}{5}$

$2 t + 0 \le - 4$

$2 t \le - 4$

$\frac{2 t}{\textcolor{red}{2}} \le - \frac{4}{\textcolor{red}{2}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} t}{\cancel{\textcolor{red}{2}}} \le - 2$

$t \le - 2$

**The Solution Is:

$t < - 2$; $t \ge 5$

Or, in interval notation:

(-oo, -2]; [5, +oo)