Solving Equations with Fractions Worksheets - Math Monks - Free Printable

Problem: Solve the given equations involving fractions with variables on both sides.

We will solve each equation step by step.

---

#### Equation 1:
\[
\frac{x - 1}{5} = \frac{7}{x - 1}
\]

1. Cross-multiply to eliminate the fractions:
\[
(x - 1)(x - 1) = 5 \cdot 7
\]
\[
(x - 1)^2 = 35
\]

2. Take the square root of both sides:
\[
x - 1 = \pm \sqrt{35}
\]

3. Solve for \( x \):
\[
x = 1 + \sqrt{35} \quad \text{or} \quad x = 1 - \sqrt{35}
\]

4. Check for extraneous solutions: The original equation has \( x - 1 \) in the denominator, so \( x \neq 1 \). Both solutions \( x = 1 + \sqrt{35} \) and \( x = 1 - \sqrt{35} \) are valid.

Solution:
\[
\boxed{x = 1 + \sqrt{35}, \, x = 1 - \sqrt{35}}
\]

---

#### Equation 2:
\[
\frac{2}{x} = 1 - \frac{3}{x + 2}
\]

1. Find a common denominator for the right-hand side:
\[
\frac{2}{x} = \frac{x + 2}{x + 2} - \frac{3}{x + 2}
\]
\[
\frac{2}{x} = \frac{x + 2 - 3}{x + 2}
\]
\[
\frac{2}{x} = \frac{x - 1}{x + 2}
\]

2. Cross-multiply to eliminate the fractions:
\[
2(x + 2) = x(x - 1)
\]
\[
2x + 4 = x^2 - x
\]

3. Rearrange into standard quadratic form:
\[
x^2 - x - 2x - 4 = 0
\]
\[
x^2 - 3x - 4 = 0
\]

4. Factor the quadratic equation:
\[
(x - 4)(x + 1) = 0
\]

5. Solve for \( x \):
\[
x = 4 \quad \text{or} \quad x = -1
\]

6. Check for extraneous solutions: The original equation has denominators \( x \) and \( x + 2 \), so \( x \neq 0 \) and \( x \neq -2 \). Both solutions \( x = 4 \) and \( x = -1 \) are valid.

Solution:
\[
\boxed{x = 4, \, x = -1}
\]

---

#### Equation 3:
\[
\frac{2x - 1}{8} = \frac{3}{4x} + 9
\]

1. Eliminate the fraction on the left-hand side by multiplying through by 8:
\[
2x - 1 = \frac{24}{4x} + 72
\]
\[
2x - 1 = \frac{6}{x} + 72
\]

2. Eliminate the fraction on the right-hand side by multiplying through by \( x \):
\[
x(2x - 1) = 6 + 72x
\]
\[
2x^2 - x = 6 + 72x
\]

3. Rearrange into standard quadratic form:
\[
2x^2 - x - 72x - 6 = 0
\]
\[
2x^2 - 73x - 6 = 0
\]

4. Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = -73 \), and \( c = -6 \):
\[
x = \frac{-(-73) \pm \sqrt{(-73)^2 - 4 \cdot 2 \cdot (-6)}}{2 \cdot 2}
\]
\[
x = \frac{73 \pm \sqrt{5329 + 48}}{4}
\]
\[
x = \frac{73 \pm \sqrt{5377}}{4}
\]

5. Simplify:
\[
x = \frac{73 + \sqrt{5377}}{4} \quad \text{or} \quad x = \frac{73 - \sqrt{5377}}{4}
\]

Solution:
\[
\boxed{x = \frac{73 + \sqrt{5377}}{4}, \, x = \frac{73 - \sqrt{5377}}{4}}
\]

---

#### Equation 4:
\[
\frac{x}{5} + \frac{2}{3} - 2 = x - \frac{2}{5}
\]

1. Simplify the constants:
\[
\frac{x}{5} + \frac{2}{3} - \frac{6}{3} = x - \frac{2}{5}
\]
\[
\frac{x}{5} - \frac{4}{3} = x - \frac{2}{5}
\]

2. Eliminate the fractions by finding a common denominator (15):
\[
\frac{3x}{15} - \frac{20}{15} = \frac{15x}{15} - \frac{6}{15}
\]
\[
3x - 20 = 15x - 6
\]

3. Rearrange terms:
\[
3x - 15x = -6 + 20
\]
\[
-12x = 14
\]

4. Solve for \( x \):
\[
x = -\frac{14}{12} = -\frac{7}{6}
\]

Solution:
\[
\boxed{x = -\frac{7}{6}}
\]

---

#### Equation 5:
\[
\frac{2}{5}x + x = -3 + \frac{1}{2}x
\]

1. Combine like terms on the left-hand side:
\[
\left(\frac{2}{5} + 1\right)x = -3 + \frac{1}{2}x
\]
\[
\left(\frac{2}{5} + \frac{5}{5}\right)x = -3 + \frac{1}{2}x
\]
\[
\frac{7}{5}x = -3 + \frac{1}{2}x
\]

2. Eliminate the fractions by multiplying through by 10:
\[
10 \cdot \frac{7}{5}x = 10 \cdot (-3) + 10 \cdot \frac{1}{2}x
\]
\[
14x = -30 + 5x
\]

3. Rearrange terms:
\[
14x - 5x = -30
\]
\[
9x = -30
\]

4. Solve for \( x \):
\[
x = -\frac{30}{9} = -\frac{10}{3}
\]

Solution:
\[
\boxed{x = -\frac{10}{3}}
\]

---

#### Equation 6:
\[
x + \frac{2}{3} = \frac{1}{4}x - 1
\]

1. Eliminate the fractions by multiplying through by 12:
\[
12 \cdot x + 12 \cdot \frac{2}{3} = 12 \cdot \frac{1}{4}x - 12 \cdot 1
\]
\[
12x + 8 = 3x - 12
\]

2. Rearrange terms:
\[
12x - 3x = -12 - 8
\]
\[
9x = -20
\]

3. Solve for \( x \):
\[
x = -\frac{20}{9}
\]

Solution:
\[
\boxed{x = -\frac{20}{9}}
\]

---

#### Equation 7:
\[
3 - \frac{1}{4}x = \frac{1}{3} + x
\]

1. Eliminate the fractions by multiplying through by 12:
\[
12 \cdot 3 - 12 \cdot \frac{1}{4}x = 12 \cdot \frac{1}{3} + 12 \cdot x
\]
\[
36 - 3x = 4 + 12x
\]

2. Rearrange terms:
\[
36 - 4 = 12x + 3x
\]
\[
32 = 15x
\]

3. Solve for \( x \):
\[
x = \frac{32}{15}
\]

Solution:
\[
\boxed{x = \frac{32}{15}}
\]

---

#### Equation 8:
\[
\frac{-4(x + 5)}{7} = \frac{12x}{7(x - 2)}
\]

1. Eliminate the denominators by multiplying through by \( 7(x - 2) \):
\[
7(x - 2) \cdot \frac{-4(x + 5)}{7} = 7(x - 2) \cdot \frac{12x}{7(x - 2)}
\]
\[
(x - 2)(-4(x + 5)) = 12x
\]
\[
-4(x - 2)(x + 5) = 12x
\]

2. Expand the left-hand side:
\[
-4(x^2 + 5x - 2x - 10) = 12x
\]
\[
-4(x^2 + 3x - 10) = 12x
\]
\[
-4x^2 - 12x + 40 = 12x
\]

3. Rearrange into standard quadratic form:
\[
-4x^2 - 12x - 12x + 40 = 0
\]
\[
-4x^2 - 24x + 40 = 0
\]
\[
x^2 + 6x - 10 = 0
\]

4. Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 6 \), and \( c = -10 \):
\[
x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1}
\]
\[
x = \frac{-6 \pm \sqrt{36 + 40}}{2}
\]
\[
x = \frac{-6 \pm \sqrt{76}}{2}
\]
\[
x = \frac{-6 \pm 2\sqrt{19}}{2}
\]
\[
x = -3 \pm \sqrt{19}
\]

5. Check for extraneous solutions: The original equation has denominators \( 7 \) and \( 7(x - 2) \), so \( x \neq 2 \). Both solutions \( x = -3 + \sqrt{19} \) and \( x = -3 - \sqrt{19} \) are valid.

Solution:
\[
\boxed{x = -3 + \sqrt{19}, \, x = -3 - \sqrt{19}}
\]

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Final Answers:

1. \(\boxed{x = 1 + \sqrt{35}, \, x = 1 - \sqrt{35}}\)
2. \(\boxed{x = 4, \, x = -1}\)
3. \(\boxed{x = \frac{73 + \sqrt{5377}}{4}, \, x = \frac{73 - \sqrt{5377}}{4}}\)
4. \(\boxed{x = -\frac{7}{6}}\)
5. \(\boxed{x = -\frac{10}{3}}\)
6. \(\boxed{x = -\frac{20}{9}}\)
7. \(\boxed{x = \frac{32}{15}}\)
8. \(\boxed{x = -3 + \sqrt{19}, \, x = -3 - \sqrt{19}}\)