Seventh Grade Two-Step Inequalities Activity (teacher made) - Free Printable

Looking at the image, I can see two worksheets titled "Two-Step Inequalities." Each worksheet has 12 problems (numbered 1 through 12), and for each problem, there are four numbers listed below it. The instruction says: “Solve each inequality and circle the values that would make the inequality true.”

Since the user didn’t specify which option to use (ExplainAnswer, TeachTopic, or GeneratePractice), and all three options are set to false in their request, I will proceed directly to solving one representative problem from the worksheet as an example — but since no specific problem was selected by the user, I’ll solve Problem #1 from the right-hand sheet as a demonstration.

Problem #1 on the right sheet is:
> ½x + 8 > 2
> Options: -36, 2, 36, -48

Let’s solve this step by step.

Step 1: Subtract 8 from both sides to isolate the term with x.
½x + 8 – 8 > 2 – 8
→ ½x > -6

Step 2: Multiply both sides by 2 to solve for x.
(½x) × 2 > -6 × 2
→ x > -12

Now check which of the given numbers are greater than -12:
- -36 → No, because -36 < -12
- 2 → Yes, because 2 > -12
- 36 → Yes, because 36 > -12
- -48 → No, because -48 < -12

So the correct answers to circle are: 2 and 36.

But wait — the user didn’t ask me to solve any particular problem. They just uploaded the image. Since they said “solve the problem accurately” but didn’t pick one, and all options are false, perhaps they expect me to recognize that without a specific question, I can’t give a single final answer.

However, looking again at the rules: “CRITICAL: First work through the problem step by step... ONLY THEN provide the final answer.” And “If you're solving math or logic problems, verify your answer before writing it in the Final Answer section.”

Since no specific problem was chosen, maybe I should solve all 12? That seems excessive. Alternatively, perhaps the intent is for me to demonstrate understanding by solving one clearly.

Given the ambiguity, I’ll choose Problem #1 from the left sheet as well to show consistency.

Left Sheet, Problem #1:
> ½y + 8 > 2
> Options: -36, 2, 36, -48

This is identical to the right sheet’s Problem #1! So same solution: y > -12 → circle 2 and 36.

Actually, let me double-check if these are truly duplicates. Looking closely:

Right Sheet Problem #1: ½x + 8 > 2 → options: -36, 2, 36, -48
Left Sheet Problem #1: ½y + 8 > 2 → options: -36, 2, 36, -48

Yes, exactly the same structure. Probably intentional for practice.

To be thorough, let’s solve another one quickly — say Right Sheet Problem #2:

> 10 - x ≤ 7
> Options: -3, 22, 3, 41

Step 1: Subtract 10 from both sides.
10 - x - 10 ≤ 7 - 10
→ -x ≤ -3

Step 2: Multiply both sides by -1. Remember: when multiplying/dividing by negative, flip the inequality sign.
(-x)(-1) ≥ (-3)(-1)
→ x ≥ 3

Check options:
- -3 → No, -3 < 3
- 22 → Yes, 22 ≥ 3
- 3 → Yes, 3 ≥ 3 (since it's "greater than or equal")
- 41 → Yes, 41 ≥ 3

So circle: 22, 3, 41

Wait — but the original instruction says “circle the values that would make the inequality true.” So multiple answers may be correct per problem.

Since the user hasn't specified which problem to solve, and given the constraints, I think the best approach is to select one clear problem and solve it completely, then present that as the final answer.

I'll go with Left Sheet, Problem #1: ½y + 8 > 2, options: -36, 2, 36, -48

As solved earlier:
½y + 8 > 2
Subtract 8: ½y > -6
Multiply by 2: y > -12

Values > -12: 2 and 36

Final verification:
Plug in y = 2: ½(2) + 8 = 1 + 8 = 9 > 2 → True
Plug in y = 36: ½(36) + 8 = 18 + 8 = 26 > 2 → True
Plug in y = -36: ½(-36) + 8 = -18 + 8 = -10 > 2? False
Plug in y = -48: ½(-48) + 8 = -24 + 8 = -16 > 2? False

Perfect.

Final Answer:
2, 36