Free Multi-step Inequalities Worksheets for Students - Free Printable

Let’s solve each of the 8 multi-step inequalities with fractions one by one, showing all steps and explaining the reasoning. Since we can’t graph here, I’ll describe how to graph each solution on a number line.

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①  $\frac{3}{5}(2d - 5) \leq 4\left(7 - \frac{1}{5}d\right)$

Step 1: Distribute both sides

Left: $\frac{3}{5} \cdot 2d = \frac{6}{5}d$, $\frac{3}{5} \cdot (-5) = -3$
→ Left side: $\frac{6}{5}d - 3$

Right: $4 \cdot 7 = 28$, $4 \cdot (-\frac{1}{5}d) = -\frac{4}{5}d$
→ Right side: $28 - \frac{4}{5}d$

Inequality becomes:
$$
\frac{6}{5}d - 3 \leq 28 - \frac{4}{5}d
$$

Step 2: Eliminate fractions by multiplying entire inequality by 5

$$
5 \cdot \left(\frac{6}{5}d - 3\right) \leq 5 \cdot \left(28 - \frac{4}{5}d\right)
$$
→ $6d - 15 \leq 140 - 4d$

Step 3: Move variables to left, constants to right

Add $4d$ to both sides:
→ $10d - 15 \leq 140$

Add 15 to both sides:
→ $10d \leq 155$

Step 4: Divide by 10

→ $d \leq \frac{155}{10} = \boxed{\frac{31}{2}}$ or $\boxed{15.5}$

✔ Solution: $d \leq \frac{31}{2}$
📌 Graph: Closed circle at 15.5, shaded to the left.

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②  $\frac{6d}{5} - 3 + 3 \leq 28 - \frac{4d}{5} + 3$

Simplify both sides first:

Left: $-3 + 3 = 0$ → $\frac{6d}{5}$
Right: $28 + 3 = 31$ → $31 - \frac{4d}{5}$

So:
$$
\frac{6d}{5} \leq 31 - \frac{4d}{5}
$$

Step 1: Multiply both sides by 5 to eliminate denominators

→ $6d \leq 155 - 4d$

Step 2: Add 4d to both sides

→ $10d \leq 155$

Step 3: Divide by 10

→ $d \leq \frac{155}{10} = \boxed{\frac{31}{2}}$ or $\boxed{15.5}$

✔ Solution: $d \leq \frac{31}{2}$
📌 Graph: Same as #1 — closed circle at 15.5, shaded left.

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③  $\frac{2x - 3}{4} + 9 \geq 3 + \frac{4x}{3}$

Step 1: Subtract 9 from both sides

→ $\frac{2x - 3}{4} \geq -6 + \frac{4x}{3}$

Step 2: Multiply entire inequality by 12 (LCM of 4 and 3)

→ $12 \cdot \frac{2x - 3}{4} \geq 12 \cdot \left(-6 + \frac{4x}{3}\right)$

→ $3(2x - 3) \geq -72 + 16x$

→ $6x - 9 \geq -72 + 16x$

Step 3: Move variables to right, constants to left

Subtract $6x$ from both sides:
→ $-9 \geq -72 + 10x$

Add 72 to both sides:
→ $63 \geq 10x$

Step 4: Divide by 10

→ $\frac{63}{10} \geq x$ → $\boxed{x \leq \frac{63}{10}}$ or $\boxed{6.3}$

✔ Solution: $x \leq \frac{63}{10}$
📌 Graph: Closed circle at 6.3, shaded left.

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④  $\frac{3x + 1}{x - 1} - 2 \geq 0$

This is a rational inequality. We must be careful about the domain: denominator ≠ 0 → $x \ne 1$

Step 1: Combine terms into single fraction

Write 2 as $\frac{2(x - 1)}{x - 1}$:

$$
\frac{3x + 1}{x - 1} - \frac{2(x - 1)}{x - 1} \geq 0
$$

→ $\frac{3x + 1 - 2x + 2}{x - 1} \geq 0$

→ $\frac{x + 3}{x - 1} \geq 0$

Step 2: Find critical points

Numerator zero: $x + 3 = 0$ → $x = -3$
Denominator zero: $x = 1$ (excluded)

Step 3: Test intervals

Intervals: $(-\infty, -3)$, $(-3, 1)$, $(1, \infty)$

- Test $x = -4$: $\frac{-4+3}{-4-1} = \frac{-1}{-5} = 0.2 > 0$ ✔
- Test $x = 0$: $\frac{3}{-1} = -3 < 0$ ✘
- Test $x = 2$: $\frac{5}{1} = 5 > 0$ ✔

Also, check endpoints: numerator zero at $x = -3$ → included (≥), denominator undefined at $x=1$ → excluded.

✔ Solution: $x \leq -3$ or $x > 1$
In interval notation: $(-\infty, -3] \cup (1, \infty)$

📌 Graph: Closed circle at -3, open circle at 1; shade left of -3 and right of 1.

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⑤  $2\left(\frac{2x + 2}{4} - \frac{3x - 3}{6}\right) < 2x - 4$

Step 1: Simplify inside parentheses

First, simplify each fraction:

$\frac{2x + 2}{4} = \frac{2(x + 1)}{4} = \frac{x + 1}{2}$

$\frac{3x - 3}{6} = \frac{3(x - 1)}{6} = \frac{x - 1}{2}$

So inside: $\frac{x + 1}{2} - \frac{x - 1}{2} = \frac{(x + 1) - (x - 1)}{2} = \frac{2}{2} = 1$

Now multiply by 2: $2 \cdot 1 = 2$

So inequality becomes:
$$
2 < 2x - 4
$$

Step 2: Solve

Add 4 to both sides:
→ $6 < 2x$

Divide by 2:
→ $3 < x$ → $\boxed{x > 3}$

✔ Solution: $x > 3$
📌 Graph: Open circle at 3, shaded right.

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⑥  $-9 > -\frac{1}{3}x + 6$

Step 1: Subtract 6 from both sides

→ $-15 > -\frac{1}{3}x$

Step 2: Multiply both sides by -3 (flip inequality!)

→ $45 < x$ → $\boxed{x > 45}$

✔ Solution: $x > 45$
📌 Graph: Open circle at 45, shaded right.

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⑦  $\frac{3x + 4}{3} - \frac{15x}{3} < 6$

Since denominators are same, combine numerators:

→ $\frac{3x + 4 - 15x}{3} < 6$

→ $\frac{-12x + 4}{3} < 6$

Step 1: Multiply both sides by 3

→ $-12x + 4 < 18$

Step 2: Subtract 4

→ $-12x < 14$

Step 3: Divide by -12 (flip inequality!)

→ $x > \frac{14}{-12} = \boxed{-\frac{7}{6}}$

✔ Solution: $x > -\frac{7}{6}$
📌 Graph: Open circle at $-\frac{7}{6}$, shaded right.

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⑧  $\frac{3}{5}x - 3 \geq \frac{3}{10}x - 9$

Step 1: Multiply entire inequality by 10 to eliminate denominators

→ $10 \cdot \left(\frac{3}{5}x - 3\right) \geq 10 \cdot \left(\frac{3}{10}x - 9\right)$

→ $6x - 30 \geq 3x - 90$

Step 2: Subtract 3x from both sides

→ $3x - 30 \geq -90$

Step 3: Add 30 to both sides

→ $3x \geq -60$

Step 4: Divide by 3

→ $x \geq \boxed{-20}$

✔ Solution: $x \geq -20$
📌 Graph: Closed circle at -20, shaded right.

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## ✔ Final Answers Summary:

| Problem | Solution |
|---------|----------|
| ① | $d \leq \frac{31}{2}$ |
| ② | $d \leq \frac{31}{2}$ |
| ③ | $x \leq \frac{63}{10}$ |
| ④ | $x \leq -3$ or $x > 1$ |
| ⑤ | $x > 3$ |
| ⑥ | $x > 45$ |
| ⑦ | $x > -\frac{7}{6}$ |
| ⑧ | $x \geq -20$ |

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Let me know if you’d like visual graphs for any of these!