Question

mbers 7-8, solve the compound inequalities
$-3m + 6 < -3$ or $4m - 3 < 5$

Explanation:

Step1: Solve \(-3m + 6 < -3\)

Subtract 6 from both sides: \(-3m + 6 - 6 < -3 - 6\)
Simplify: \(-3m < -9\)
Divide both sides by -3 (reverse inequality): \(m > 3\)

Step2: Solve \(4m - 3 < 5\)

Add 3 to both sides: \(4m - 3 + 3 < 5 + 3\)
Simplify: \(4m < 8\)
Divide by 4: \(m < 2\)

Step3: Combine solutions (OR)

The compound inequality uses "or", so the solution is the union of \(m > 3\) and \(m < 2\).

Answer:

All real numbers \(m\) such that \(m < 2\) or \(m > 3\) (or in interval notation: \((-\infty, 2) \cup (3, \infty)\))