Question

solve for a and graph the solution.
11 ≥ 3a + 20 ≥ 2
plot the endpoints. select an endpoint to change it from closed to open. select the middle of a segment, ray, or line to delete it.

Explanation:

Step1: Subtract 20 from all parts

$$11 - 20\geq3a+20 - 20\geq2 - 20$$
$$-9\geq3a\geq - 18$$

Step2: Divide all parts by 3

$$\frac{-9}{3}\geq\frac{3a}{3}\geq\frac{-18}{3}$$
$$- 3\geq a\geq - 6$$

Answer:

The solution is \(-6\leq a\leq - 3\). On the number - line, we plot closed circles at \(-6\) and \(-3\) (since the inequalities are non - strict, i.e., \(\geq\) and \(\leq\)) and draw a line segment connecting them.