Question

1. solve the compound inequality: $2a + 10 \leq 6$ or $3 + 3a > 3$

a. $a > 1$
b. $a < -9$
c. $a \leq -2$ or $a > 0$
d. all real numbers

Explanation:

Step1: Solve \(2a + 10 \leq 6\)

Subtract 10 from both sides: \(2a + 10 - 10 \leq 6 - 10\)
Simplify: \(2a \leq -4\)
Divide both sides by 2: \(\frac{2a}{2} \leq \frac{-4}{2}\)
Result: \(a \leq -2\)

Step2: Solve \(3 + 3a > 3\)

Subtract 3 from both sides: \(3 + 3a - 3 > 3 - 3\)
Simplify: \(3a > 0\)
Divide both sides by 3: \(\frac{3a}{3} > \frac{0}{3}\)
Result: \(a > 0\)

Step3: Combine the solutions (OR)

The compound inequality is "or", so we take the union of \(a \leq -2\) and \(a > 0\). Wait, but let's check the options. Wait, maybe I made a mistake? Wait, no, let's re - check. Wait, the options: Let's see, option C is \(a \leq - 2\) or \(a > 0\), but wait, let's check the solving again. Wait, for \(2a+10\leq6\): \(2a\leq6 - 10=-4\), \(a\leq - 2\). For \(3 + 3a>3\): \(3a>0\), \(a > 0\). So the solution is \(a\leq - 2\) or \(a > 0\), which is option C? Wait, but wait, let's check the options again. Wait, the options are:

A. \(a>1\)

B. \(a < - 9\)

C. \(a\leq - 2\) or \(a>0\)

D. all real numbers

Wait, but when we solve \(2a + 10\leq6\), we get \(a\leq - 2\), and when we solve \(3+3a > 3\), we get \(a>0\). The "or" in the compound inequality means that the solution is all numbers that satisfy either of the inequalities. So the solution is \(a\leq - 2\) or \(a > 0\), which is option C. Wait, but let's check again. Wait, maybe I miscalculated? Let's re - solve \(3 + 3a>3\):

\(3+3a>3\)

Subtract 3 from both sides: \(3a>0\), so \(a > 0\). Correct.

\(2a + 10\leq6\):

\(2a\leq6 - 10=-4\), \(a\leq - 2\). Correct.

So the solution is \(a\leq - 2\) or \(a > 0\), which is option C.

Answer:

C. \(a\leq - 2\) or \(a>0\)