Question

1. which inequality is solved by subtracting 5 from both sides and then dividing by 2?

a. $2x - 5 = 15$
b. $2x - 5 > 15$
c. $2x + 5 > 15$
d. $3x + 5 = 15$

Explanation:

Step1: Analyze required operations

We need an inequality solved by first subtracting 5, then dividing by 2.

Step2: Reverse-engineer the inequality

Start with the final step logic: If we divide by 2 last, the variable term must be $2x$. If we subtract 5 first, the constant term is $+5$ on the same side as $2x$.
Check option c: $2x + 5 > 15$

1. Subtract 5 from both sides: $2x + 5 - 5 > 15 - 5$ → $2x > 10$
2. Divide both sides by 2: $\frac{2x}{2} > \frac{10}{2}$ → $x > 5$

This matches the required steps. Other options do not: a/d are equations, b requires adding 5 first.

Answer:

c. $2x + 5 > 15$