Answer:
[tex]\{a[/tex] ∈ [tex]R\ -21 \le a \le \infty \}[/tex] --- set builder
[tex][-21,\infty)[/tex] --- interval notation
Step-by-step explanation:
Given
[tex]-3a - 15 \le -2a + 6[/tex]
Required
Solve
Collect like terms
[tex]-3a + 2a \le 15 + 6[/tex]
[tex]-a \le 21[/tex]
Divide by -1
[tex]a \ge - 21[/tex]
Rewrite as:
[tex]-21 \le a[/tex]
Using set builder
[tex]\{a[/tex] ∈ [tex]R\ -21 \le a \le \infty \}[/tex]
Using interval notation, we have:
[tex][-21,\infty)[/tex]