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Math worksheet providing eight practice problems for solving multi-step equations with fractions.

Algebra worksheet featuring 8 multi-step equations with fractions for students to solve.

Algebra worksheet featuring 8 multi-step equations with fractions for students to solve.

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Show Answer Key & Explanations Step-by-step solution for: Multi-Step Equations with Fractions Worksheets
Let's solve each of these multi-step equations involving fractions step by step. I'll go through each one carefully, showing the algebraic steps and simplifications.

---

1) $ 5\left(2h + \frac{1}{3}\right) = 7h + 11 $



Step 1: Distribute the 5 on the left side:
$$
5 \cdot 2h + 5 \cdot \frac{1}{3} = 7h + 11
\Rightarrow 10h + \frac{5}{3} = 7h + 11
$$

Step 2: Subtract $7h$ from both sides:
$$
10h - 7h + \frac{5}{3} = 11 \Rightarrow 3h + \frac{5}{3} = 11
$$

Step 3: Subtract $\frac{5}{3}$ from both sides:
$$
3h = 11 - \frac{5}{3} = \frac{33}{3} - \frac{5}{3} = \frac{28}{3}
$$

Step 4: Divide both sides by 3:
$$
h = \frac{28}{3} \div 3 = \frac{28}{9}
$$

Answer: $ h = \boxed{\frac{28}{9}} $

---

2) $ \frac{9}{7}(1 + x) - \frac{6}{7}x + 1 = 1 $



Step 1: Distribute $ \frac{9}{7} $:
$$
\frac{9}{7} + \frac{9}{7}x - \frac{6}{7}x + 1 = 1
$$

Step 2: Combine like terms:
- $ \frac{9}{7}x - \frac{6}{7}x = \frac{3}{7}x $
- Constants: $ \frac{9}{7} + 1 = \frac{9}{7} + \frac{7}{7} = \frac{16}{7} $

So:
$$
\frac{3}{7}x + \frac{16}{7} = 1
$$

Step 3: Subtract $ \frac{16}{7} $ from both sides:
$$
\frac{3}{7}x = 1 - \frac{16}{7} = \frac{7}{7} - \frac{16}{7} = -\frac{9}{7}
$$

Step 4: Multiply both sides by $ \frac{7}{3} $:
$$
x = -\frac{9}{7} \cdot \frac{7}{3} = -\frac{9}{3} = -3
$$

Answer: $ x = \boxed{-3} $

---

3) $ \frac{7}{8}t - \frac{4}{7}t = \frac{5}{6}t - \frac{3}{4} $



Step 1: Get all terms with $ t $ on one side. Subtract $ \frac{5}{6}t $ from both sides:
$$
\frac{7}{8}t - \frac{4}{7}t - \frac{5}{6}t = -\frac{3}{4}
$$

Step 2: Find a common denominator for the coefficients (LCM of 8, 7, 6 is 168):
- $ \frac{7}{8} = \frac{147}{168} $
- $ \frac{4}{7} = \frac{96}{168} $
- $ \frac{5}{6} = \frac{140}{168} $

Now:
$$
\frac{147}{168}t - \frac{96}{168}t - \frac{140}{168}t = -\frac{3}{4}
\Rightarrow \left(\frac{147 - 96 - 140}{168}\right)t = -\frac{3}{4}
\Rightarrow \frac{-89}{168}t = -\frac{3}{4}
$$

Step 3: Divide both sides by $ -\frac{89}{168} $ (or multiply by reciprocal):
$$
t = -\frac{3}{4} \div \left(-\frac{89}{168}\right) = \frac{3}{4} \cdot \frac{168}{89}
$$

Simplify:
- $ \frac{3}{4} \cdot \frac{168}{89} = \frac{3 \cdot 168}{4 \cdot 89} = \frac{504}{356} $

Reduce: divide numerator and denominator by 4:
- $ \frac{504 \div 4}{356 \div 4} = \frac{126}{89} $

Answer: $ t = \boxed{\frac{126}{89}} $

---

4) $ \frac{8}{5} + 3z = 10z - 4 $



Step 1: Move all $ z $-terms to one side and constants to the other.

Subtract $ 3z $ from both sides:
$$
\frac{8}{5} = 7z - 4
$$

Add 4 to both sides:
$$
\frac{8}{5} + 4 = 7z \Rightarrow \frac{8}{5} + \frac{20}{5} = \frac{28}{5} = 7z
$$

Step 2: Divide both sides by 7:
$$
z = \frac{28}{5} \div 7 = \frac{28}{5} \cdot \frac{1}{7} = \frac{4}{5}
$$

Answer: $ z = \boxed{\frac{4}{5}} $

---

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



Step 1: Subtract $ \frac{3}{2}h $ from both sides:
$$
-10 = 4h - \frac{3}{2}h = \left(\frac{8}{2} - \frac{3}{2}\right)h = \frac{5}{2}h
$$

Step 2: Divide both sides by $ \frac{5}{2} $:
$$
h = -10 \div \frac{5}{2} = -10 \cdot \frac{2}{5} = -4
$$

Answer: $ h = \boxed{-4} $

---

6) $ \frac{3 + 4n}{4} = \frac{8}{n} $



Step 1: Eliminate denominators by multiplying both sides by $ 4n $ (LCD of 4 and $ n $):

$$
4n \cdot \left(\frac{3 + 4n}{4}\right) = 4n \cdot \left(\frac{8}{n}\right)
\Rightarrow n(3 + 4n) = 32
$$

Step 2: Expand:
$$
3n + 4n^2 = 32
\Rightarrow 4n^2 + 3n - 32 = 0
$$

Step 3: Solve quadratic equation using quadratic formula:
$$
n = \frac{-3 \pm \sqrt{3^2 - 4(4)(-32)}}{2(4)} = \frac{-3 \pm \sqrt{9 + 512}}{8} = \frac{-3 \pm \sqrt{521}}{8}
$$

Since $ \sqrt{521} $ is irrational, we leave it as is.

But let’s check if it factors:
Try factoring $ 4n^2 + 3n - 32 $. Try pairs:
- $ (4n - 13)(n + 2.5) $? Not integers.

No rational roots — so answer is:
$$
n = \frac{-3 \pm \sqrt{521}}{8}
$$

But wait — let's double-check the original equation:
$$
\frac{3 + 4n}{4} = \frac{8}{n}
$$

We can cross-multiply:
$$
n(3 + 4n) = 32 \Rightarrow 4n^2 + 3n - 32 = 0
$$

Yes, correct.

Answer: $ n = \boxed{\frac{-3 \pm \sqrt{521}}{8}} $

---

7) $ \frac{u - 5}{8} = 2u - 1 $



Step 1: Multiply both sides by 8 to eliminate denominator:
$$
u - 5 = 8(2u - 1) = 16u - 8
$$

Step 2: Move all terms to one side:
$$
u - 5 - 16u + 8 = 0 \Rightarrow -15u + 3 = 0
\Rightarrow -15u = -3
\Rightarrow u = \frac{-3}{-15} = \frac{1}{5}
$$

Answer: $ u = \boxed{\frac{1}{5}} $

---

8) $ 6v + \frac{4}{9} = 4\left(\frac{1}{2} + v\right) $



Step 1: Distribute the 4 on the right:
$$
6v + \frac{4}{9} = 4 \cdot \frac{1}{2} + 4v = 2 + 4v
$$

Step 2: Subtract $ 4v $ from both sides:
$$
2v + \frac{4}{9} = 2
$$

Step 3: Subtract $ \frac{4}{9} $ from both sides:
$$
2v = 2 - \frac{4}{9} = \frac{18}{9} - \frac{4}{9} = \frac{14}{9}
$$

Step 4: Divide by 2:
$$
v = \frac{14}{9} \div 2 = \frac{14}{9} \cdot \frac{1}{2} = \frac{7}{9}
$$

Answer: $ v = \boxed{\frac{7}{9}} $

---

Final Answers:



1) $ h = \boxed{\frac{28}{9}} $
2) $ x = \boxed{-3} $
3) $ t = \boxed{\frac{126}{89}} $
4) $ z = \boxed{\frac{4}{5}} $
5) $ h = \boxed{-4} $
6) $ n = \boxed{\frac{-3 \pm \sqrt{521}}{8}} $
7) $ u = \boxed{\frac{1}{5}} $
8) $ v = \boxed{\frac{7}{9}} $

Let me know if you'd like these checked or explained in more detail!
Parent Tip: Review the logic above to help your child master the concept of 8th grade fraction worksheet.
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