Solve the inequality: 3(x + 1) $\leq$ 5(x + 2) + 15

Analyze the question: We need to find all values of x that satisfy 3(x + 1) ≤ 5(x + 2) + 15.

Concept: Solve an inequality like an equation: expand brackets, collect like terms, and then divide. The key rule is that when you multiply or divide by a negative number, the inequality sign flips.

Step-by-step solution:

1. Expand both sides: 3x + 3 ≤ 5x + 10 + 15 3x + 3 ≤ 5x + 25

2. Move the x terms to one side and numbers to the other: 3 − 25 ≤ 5x − 3x −22 ≤ 2x

3. Divide both sides by 2 (a positive number, so the sign stays the same): −11 ≤ x

This is the same as x ≥ −11.

Check: Let x = −11:

• Left side: 3(−11 + 1) = 3(−10) = −30
• Right side: 5(−11 + 2) + 15 = 5(−9) + 15 = −45 + 15 = −30

Both sides are equal at x = −11, so it is the boundary. Try x = 0: left = 3, right = 25, and 3 ≤ 25 is true. So values greater than −11 work.

Answer: x ≥ −11, which is option C.

Tip: Be careful to read "−11 ≤ x" correctly. The inequality sign "opens" toward the larger value, so x is the larger one, meaning x ≥ −11. Also, always test a simple value such as x = 0 to confirm your direction.