Grade 6 Complex Fractions Worksheets 2024 - Free Printable

Let's solve each of these multi-step equations involving fractions step by step. I'll go through each problem carefully, showing the steps and simplifying as we go.

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1) $ \frac{5}{2}\left(2h + \frac{4}{5}\right) = 7h + 11 $

Step 1: Distribute $ \frac{5}{2} $ on the left side:

$$
\frac{5}{2} \cdot 2h + \frac{5}{2} \cdot \frac{4}{5} = 7h + 11
$$

$$
5h + 2 = 7h + 11
$$

Step 2: Subtract $ 5h $ from both sides:

$$
2 = 2h + 11
$$

Step 3: Subtract 11 from both sides:

$$
-9 = 2h
$$

Step 4: Divide by 2:

$$
h = -\frac{9}{2}
$$

✔ Answer: $ h = -\frac{9}{2} $

---

2) $ \frac{3}{2}(1 + x) - \frac{6}{5}x + 1 = 1 $

Step 1: Distribute $ \frac{3}{2} $:

$$
\frac{3}{2} + \frac{3}{2}x - \frac{6}{5}x + 1 = 1
$$

Step 2: Combine like terms (constants and $ x $-terms):

Constants: $ \frac{3}{2} + 1 = \frac{5}{2} $

$ x $-terms: $ \frac{3}{2}x - \frac{6}{5}x $

Find common denominator (LCM of 2 and 5 is 10):

$$
\frac{15}{10}x - \frac{12}{10}x = \frac{3}{10}x
$$

So equation becomes:

$$
\frac{3}{10}x + \frac{5}{2} = 1
$$

Step 3: Subtract $ \frac{5}{2} $ from both sides:

$$
\frac{3}{10}x = 1 - \frac{5}{2} = -\frac{3}{2}
$$

Step 4: Multiply both sides by reciprocal of $ \frac{3}{10} $, which is $ \frac{10}{3} $:

$$
x = -\frac{3}{2} \cdot \frac{10}{3} = -\frac{30}{6} = -5
$$

✔ Answer: $ x = -5 $

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3) $ \frac{2}{3}t - \frac{4}{7} - \frac{5}{6}t - \frac{2}{7} = 0 $

Step 1: Combine like terms.

Group $ t $-terms and constants:

$ t $-terms: $ \frac{2}{3}t - \frac{5}{6}t $

Common denominator: 6

$$
\frac{4}{6}t - \frac{5}{6}t = -\frac{1}{6}t
$$

Constants: $ -\frac{4}{7} - \frac{2}{7} = -\frac{6}{7} $

So equation becomes:

$$
-\frac{1}{6}t - \frac{6}{7} = 0
$$

Step 2: Add $ \frac{6}{7} $ to both sides:

$$
-\frac{1}{6}t = \frac{6}{7}
$$

Step 3: Multiply both sides by $ -6 $:

$$
t = -6 \cdot \frac{6}{7} = -\frac{36}{7}
$$

✔ Answer: $ t = -\frac{36}{7} $

---

4) $ \frac{1}{5}z + 3z = 10z - 4 $

Step 1: Combine like terms on the left side:

$ \frac{1}{5}z + 3z = \frac{1}{5}z + \frac{15}{5}z = \frac{16}{5}z $

So:

$$
\frac{16}{5}z = 10z - 4
$$

Step 2: Subtract $ 10z $ from both sides:

$$
\frac{16}{5}z - 10z = -4
$$

Convert $ 10z $ to $ \frac{50}{5}z $:

$$
\frac{16}{5}z - \frac{50}{5}z = -\frac{34}{5}z = -4
$$

Step 3: Divide both sides by $ -\frac{34}{5} $:

$$
z = -4 \div \left(-\frac{34}{5}\right) = -4 \cdot \left(-\frac{5}{34}\right) = \frac{20}{34} = \frac{10}{17}
$$

✔ Answer: $ z = \frac{10}{17} $

---

5) $ \frac{3}{2}h - 10 = 4h $

Step 1: Subtract $ \frac{3}{2}h $ from both sides:

$$
-10 = 4h - \frac{3}{2}h
$$

Convert $ 4h = \frac{8}{2}h $:

$$
\frac{8}{2}h - \frac{3}{2}h = \frac{5}{2}h
$$

So:

$$
-10 = \frac{5}{2}h
$$

Step 2: Multiply both sides by $ \frac{2}{5} $:

$$
h = -10 \cdot \frac{2}{5} = -4
$$

✔ Answer: $ h = -4 $

---

6) $ \frac{3 + 4n}{4} - \frac{6}{5} = n $

Step 1: Simplify the left side:

Break $ \frac{3 + 4n}{4} $ into $ \frac{3}{4} + \frac{4n}{4} = \frac{3}{4} + n $

So:

$$
\frac{3}{4} + n - \frac{6}{5} = n
$$

Step 2: Subtract $ n $ from both sides:

$$
\frac{3}{4} - \frac{6}{5} = 0
$$

Now compute $ \frac{3}{4} - \frac{6}{5} $:

LCM of 4 and 5 is 20:

$$
\frac{15}{20} - \frac{24}{20} = -\frac{9}{20}
$$

So:

$$
-\frac{9}{20} = 0
$$

✘ This is not true, so there’s a contradiction.

Wait — this means our assumption must be wrong? Let's double-check.

Original:

$$
\frac{3 + 4n}{4} - \frac{6}{5} = n
$$

We simplified left side: $ \frac{3}{4} + n - \frac{6}{5} = n $

Then subtract $ n $: $ \frac{3}{4} - \frac{6}{5} = 0 $ → $ -\frac{9}{20} = 0 $? No.

So no solution?

But that can’t be right — maybe we made a mistake in interpretation.

Wait: actually, let's not cancel $ n $ prematurely.

Go back:

$$
\frac{3}{4} + n - \frac{6}{5} = n
$$

Subtract $ n $ from both sides:

$$
\frac{3}{4} - \frac{6}{5} = 0
$$

→ $ -\frac{9}{20} = 0 $? False.

So no solution?

But let's check if we made an error.

Actually, this implies that the equation is never true, unless there's a typo.

But let's re-solve carefully.

Start over:

$$
\frac{3 + 4n}{4} - \frac{6}{5} = n
$$

Multiply every term by 20 (LCM of 4 and 5) to eliminate denominators:

$$
20 \cdot \left(\frac{3 + 4n}{4}\right) - 20 \cdot \frac{6}{5} = 20 \cdot n
$$

$$
5(3 + 4n) - 24 = 20n
$$

$$
15 + 20n - 24 = 20n
$$

$$
-9 + 20n = 20n
$$

Subtract $ 20n $ from both sides:

$$
-9 = 0
$$

✘ Contradiction.

So no solution.

But wait — is that possible?

Yes. The equation has no solution because it reduces to a false statement.

✔ Answer: No solution (or "No real solution")

But let's check the original problem again.

Wait — perhaps it was meant to be $ \frac{3 + 4n}{4} - \frac{6}{5} = n $. But as written, it leads to contradiction.

So unless there's a typo, no solution.

But let's assume it's correct.

✔ Answer: No solution

But let's see if maybe the equation was misread.

Wait — could it be $ \frac{3 + 4n}{4} - \frac{6}{5} = n $? Yes, and we just showed it leads to $ -9 = 0 $, impossible.

So no solution.

But let me write it clearly.

✔ Answer: No solution

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7) $ \frac{m - 5}{6} = 2m - 1 $

Step 1: Multiply both sides by 6 to eliminate denominator:

$$
m - 5 = 6(2m - 1)
$$

$$
m - 5 = 12m - 6
$$

Step 2: Subtract $ m $ from both sides:

$$
-5 = 11m - 6
$$

Step 3: Add 6 to both sides:

$$
1 = 11m
$$

Step 4: Divide by 11:

$$
m = \frac{1}{11}
$$

✔ Answer: $ m = \frac{1}{11} $

---

8) $ 6v + \frac{4}{5} = 4\left(\frac{1}{3}v + 3\right) $

Step 1: Expand the right side:

$$
6v + \frac{4}{5} = 4 \cdot \frac{1}{3}v + 4 \cdot 3 = \frac{4}{3}v + 12
$$

Step 2: Subtract $ \frac{4}{3}v $ from both sides:

$$
6v - \frac{4}{3}v + \frac{4}{5} = 12
$$

Convert $ 6v = \frac{18}{3}v $:

$$
\frac{18}{3}v - \frac{4}{3}v = \frac{14}{3}v
$$

So:

$$
\frac{14}{3}v + \frac{4}{5} = 12
$$

Step 3: Subtract $ \frac{4}{5} $:

$$
\frac{14}{3}v = 12 - \frac{4}{5} = \frac{60}{5} - \frac{4}{5} = \frac{56}{5}
$$

Step 4: Multiply both sides by $ \frac{3}{14} $:

$$
v = \frac{56}{5} \cdot \frac{3}{14} = \frac{168}{70} = \frac{24}{10} = \frac{12}{5}
$$

Simplify: $ \frac{12}{5} $

✔ Answer: $ v = \frac{12}{5} $

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✔ Final Answers:

1) $ h = -\frac{9}{2} $
2) $ x = -5 $
3) $ t = -\frac{36}{7} $
4) $ z = \frac{10}{17} $
5) $ h = -4 $
6) No solution
7) $ m = \frac{1}{11} $
8) $ v = \frac{12}{5} $

Let me know if you'd like explanations for any specific step!