4
algebra

Solve: 5(x+2)7x+45(x + 2) \leq 7x + 4

A

x3x \geq 3

B

x3x \leq 3

C

x3x \geq -3

D

x3x \leq -3

Correct Answer: A

Choice A is the correct answer. Distribute and solve.

  1. Distribute: 5x+107x+45x + 10 \leq 7x + 4
  2. Subtract 5x: 102x+410 \leq 2x + 4
  3. Subtract 4: 62x6 \leq 2x
  4. Divide by 2: 3x3 \leq x, which is equivalent to x3x \geq 3

?�� Strategic Tip: Reading 3x3 \leq x backwards as x3x \geq 3 is a crucial skill.

Choice B is incorrect because it reverses the inequality direction (x3x \leq 3). Choice C is incorrect because it uses negative 3. Choice D is incorrect because it uses negative 3 and wrong direction.