Question

solve the inequality.
\frac{3}{2}x + 3 < 2 or 3x + \frac{1}{5} \geq 1

x < -\frac{?}{\square} or x \geq \frac{\square}{\square}

Explanation:

Step1: Isolate first inequality's x term

Subtract 3 from both sides:
$\frac{3}{2}x + 3 - 3 < 2 - 3$
$\frac{3}{2}x < -1$

Step2: Solve first inequality for x

Multiply both sides by $\frac{2}{3}$:
$x < -1 \times \frac{2}{3}$
$x < -\frac{2}{3}$

Step3: Isolate second inequality's x term

Subtract $\frac{1}{5}$ from both sides:
$3x + \frac{1}{5} - \frac{1}{5} \geq 1 - \frac{1}{5}$
$3x \geq \frac{4}{5}$

Step4: Solve second inequality for x

Divide both sides by 3:
$x \geq \frac{4}{5} \times \frac{1}{3}$
$x \geq \frac{4}{15}$

Answer:

$x < -\frac{2}{3}$ or $x \geq \frac{4}{15}$