Question

graph the solution set for m - 8 ≤ -6 or $\frac{3}{4}m < -3$. select
o solution\ if applicable.

Explanation:

Step1: Solve the first inequality

Add 8 to both sides of $m - 8\leq - 6$.
$m-8 + 8\leq - 6+8$
$m\leq2$

Step2: Solve the second inequality

Multiply both sides of $\frac{3}{4}m\lt - 3$ by $\frac{4}{3}$.
$\frac{4}{3}\times\frac{3}{4}m\lt - 3\times\frac{4}{3}$
$m\lt - 4$

Step3: Graph the solution sets

The solution of $m\leq2$ includes all numbers less than or equal to 2. The solution of $m\lt - 4$ includes all numbers less than - 4. Since it is an "or" statement, the solution set is all real - numbers that satisfy either inequality. We draw a closed circle at 2 (because $m$ can equal 2) and shade to the left for $m\leq2$, and an open circle at - 4 (because $m$ cannot equal - 4) and shade to the left for $m\lt - 4$. The combined shaded region is all real numbers less than or equal to 2.

Answer:

Shade the number line to the left of 2, with a closed circle at 2.