Two-Step Inequalities Worksheets with Answer Key - Free Printable

Let’s solve each inequality one by one. We’ll isolate the variable step by step, just like solving equations — but remember: if you multiply or divide by a negative number, flip the inequality sign!

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①
\(\frac{-x}{5} + 6 > -3\)
Subtract 6 from both sides:
\(\frac{-x}{5} > -9\)
Multiply both sides by 5:
\(-x > -45\)
Now divide by -1 (and flip the sign!):
\(x < 45\)

✔ Final Answer for ①: \(x < 45\)

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②
\(-9 \leq -8 + \frac{m}{-6}\)
Add 8 to both sides:
\(-1 \leq \frac{m}{-6}\)
Multiply both sides by -6 (flip the sign!):
\(6 \geq m\) → which is same as \(m \leq 6\)

✔ Final Answer for ②: \(m \leq 6\)

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③
\(6h - 10h + 2 \geq -10\)
Combine like terms:
\(-4h + 2 \geq -10\)
Subtract 2:
\(-4h \geq -12\)
Divide by -4 (flip the sign!):
\(h \leq 3\)

✔ Final Answer for ③: \(h \leq 3\)

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④
\(-1 < \frac{-8 + m}{22}\)
Multiply both sides by 22:
\(-22 < -8 + m\)
Add 8 to both sides:
\(-14 < m\) → which is same as \(m > -14\)

✔ Final Answer for ④: \(m > -14\)

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⑤
\(143 \leq 11b + 11\)
Subtract 11:
\(132 \leq 11b\)
Divide by 11:
\(12 \leq b\) → which is same as \(b \geq 12\)

✔ Final Answer for ⑤: \(b \geq 12\)

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⑥
\(-2 < \frac{1 + p}{2}\)
Multiply both sides by 2:
\(-4 < 1 + p\)
Subtract 1:
\(-5 < p\) → which is same as \(p > -5\)

✔ Final Answer for ⑥: \(p > -5\)

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⑦
\(5(k + 2) > -5\)
Distribute the 5:
\(5k + 10 > -5\)
Subtract 10:
\(5k > -15\)
Divide by 5:
\(k > -3\)

✔ Final Answer for ⑦: \(k > -3\)

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⑧
\(-3 + \frac{9}{3} > -5\)
First simplify \(\frac{9}{3} = 3\), so:
\(-3 + 3 > -5\) → \(0 > -5\)
This is always true — no variable! So it’s an identity. But since there’s no variable to solve for, we say “all real numbers” satisfy this? Wait — actually, looking again: the problem says “solve each inequality”, and this one has no variable. That means it’s either always true or never true.

Since \(0 > -5\) is TRUE, then any value would work — but there’s no variable. Hmm. Maybe typo? Or perhaps they meant to have a variable? Let me check original:
Original: \(-3 + \frac{9}{3} > -5\) → yes, no variable. So technically, this inequality is always true. But in context of solving for a variable, maybe it's a trick question? Since no variable, we can’t solve for anything — but the statement is true.

Wait — let me double-check: Is it possible that it was supposed to be \(-3 + \frac{g}{3} > -5\)? Because otherwise, it’s not really an inequality to solve — it’s just a true statement.

But based on what’s written:
It simplifies to \(0 > -5\), which is TRUE. So if forced to write a solution set, it’s “all real numbers” — but since no variable, maybe just state it’s always true.

However, in most school contexts, if there’s no variable, and it’s true, we might say “true for all values” — but since no variable, perhaps leave as is? Actually, let’s assume it’s correct as written.

So: The inequality is always true → solution is all real numbers? But again, no variable. This is odd.

Wait — maybe I misread. Let me look again at image description:
Problem ⑧: \(-3 + \frac{9}{3} > -5\) — yes, no variable. So perhaps it’s a mistake, but we’ll go with logic.

Actually, in some curricula, they include such problems to test understanding. Since it’s always true, and no variable, we can say “the inequality is always true” — but for consistency, maybe they expect us to recognize it’s true regardless.

But let’s move on — perhaps it’s intentional. For now, I’ll note: This inequality contains no variable and evaluates to a true statement, so it holds for any input — but since no variable, technically no solution set needed. However, to match format, perhaps write “always true”.

But wait — let’s see other problems. All others have variables. Maybe it’s a typo and should be \(-3 + \frac{g}{3} > -5\)? If so, let’s solve that version too, just in case.

Assume it’s \(-3 + \frac{g}{3} > -5\):
Add 3: \(\frac{g}{3} > -2\)
Multiply by 3: \(g > -6\)

That makes more sense. Given that all other problems have variables, likely a typo in transcription. In the original image, it might be \(\frac{g}{3}\), not \(\frac{9}{3}\). Let me check user’s text:
User wrote: “⑧ -3 + 9/3 > -5” — oh, it’s written as 9/3, which is 3. So probably intentional? But that doesn’t make sense for a “solve” problem.

Alternatively, maybe it’s \(-3 + \frac{9}{g} > -5\)? No, user wrote “9/3”.

Given ambiguity, but to stay faithful to input:
If it’s truly \(-3 + \frac{9}{3} > -5\), then it’s \(0 > -5\), which is true, so no restriction — but since no variable, perhaps the answer is “all real numbers” or “true”.

But in standard math worksheets, if an inequality has no variable and is true, it’s considered satisfied for all x — but here no x. I think best to flag it, but for now, I’ll proceed assuming it’s a typo and should be \(-3 + \frac{g}{3} > -5\), giving \(g > -6\). Many similar problems use g or other letters.

Looking back at problem ⑨: uses q, ⑩ uses a, so ⑧ might be g. Likely typo in user’s transcription — they wrote “9” instead of “g”. Because 9/3 is constant.

I think it’s safe to assume it’s \(-3 + \frac{g}{3} > -5\), as that fits the pattern.

So solving that:
\(-3 + \frac{g}{3} > -5\)
Add 3: \(\frac{g}{3} > -2\)
Multiply by 3: \(g > -6\)

✔ Final Answer for ⑧ (assuming typo): \(g > -6\)

*(Note: If strictly as written, it’s always true, but that doesn’t fit the worksheet style.)*

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⑨
\(7 < 4q - 9\)
Add 9 to both sides:
\(16 < 4q\)
Divide by 4:
\(4 < q\) → which is same as \(q > 4\)

✔ Final Answer for ⑨: \(q > 4\)

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⑩
\(\frac{-9 + a}{15} > 1\)
Multiply both sides by 15:
\(-9 + a > 15\)
Add 9:
\(a > 24\)

✔ Final Answer for ⑩: \(a > 24\)

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Final Answers Summary:

① \(x < 45\)
② \(m \leq 6\)
③ \(h \leq 3\)
④ \(m > -14\)
⑤ \(b \geq 12\)
⑥ \(p > -5\)
⑦ \(k > -3\)
⑧ \(g > -6\) *(assuming typo; if not, inequality is always true)*
⑨ \(q > 4\)
⑩ \(a > 24\)

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Final Answer:
① \(x < 45\)
② \(m \leq 6\)
③ \(h \leq 3\)
④ \(m > -14\)
⑤ \(b \geq 12\)
⑥ \(p > -5\)
⑦ \(k > -3\)
⑧ \(g > -6\)
⑨ \(q > 4\)
⑩ \(a > 24\)