Practice Questions on Solving Equations with Fractions

Problem: Solve the given equations involving fractions.

We will solve each equation step by step. Let's start with Q1 and then move on to Q2.

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Q1: Solve the following equations

#### [a] \( \frac{2x + 3}{x - 1} = 7 \)

1. Eliminate the denominator: Multiply both sides by \( x - 1 \) (assuming \( x \neq 1 \)):
\[
2x + 3 = 7(x - 1)
\]

2. Expand and simplify:
\[
2x + 3 = 7x - 7
\]

3. Rearrange terms to isolate \( x \):
\[
2x - 7x = -7 - 3
\]
\[
-5x = -10
\]

4. Solve for \( x \):
\[
x = \frac{-10}{-5} = 2
\]

5. Verify the solution: Substitute \( x = 2 \) back into the original equation:
\[
\frac{2(2) + 3}{2 - 1} = \frac{4 + 3}{1} = 7
\]
The solution is correct.

Solution for [a]: \( x = 2 \)

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#### [b] \( \frac{4x - 2}{x + 5} = 2 \)

1. Eliminate the denominator: Multiply both sides by \( x + 5 \) (assuming \( x \neq -5 \)):
\[
4x - 2 = 2(x + 5)
\]

2. Expand and simplify:
\[
4x - 2 = 2x + 10
\]

3. Rearrange terms to isolate \( x \):
\[
4x - 2x = 10 + 2
\]
\[
2x = 12
\]

4. Solve for \( x \):
\[
x = \frac{12}{2} = 6
\]

5. Verify the solution: Substitute \( x = 6 \) back into the original equation:
\[
\frac{4(6) - 2}{6 + 5} = \frac{24 - 2}{11} = \frac{22}{11} = 2
\]
The solution is correct.

Solution for [b]: \( x = 6 \)

---

#### [c] \( \frac{5x - 3}{x - 2} = 4 \)

1. Eliminate the denominator: Multiply both sides by \( x - 2 \) (assuming \( x \neq 2 \)):
\[
5x - 3 = 4(x - 2)
\]

2. Expand and simplify:
\[
5x - 3 = 4x - 8
\]

3. Rearrange terms to isolate \( x \):
\[
5x - 4x = -8 + 3
\]
\[
x = -5
\]

4. Verify the solution: Substitute \( x = -5 \) back into the original equation:
\[
\frac{5(-5) - 3}{-5 - 2} = \frac{-25 - 3}{-7} = \frac{-28}{-7} = 4
\]
The solution is correct.

Solution for [c]: \( x = -5 \)

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#### [d] \( \frac{3x - 3}{x + 3} = 6 \)

1. Eliminate the denominator: Multiply both sides by \( x + 3 \) (assuming \( x \neq -3 \)):
\[
3x - 3 = 6(x + 3)
\]

2. Expand and simplify:
\[
3x - 3 = 6x + 18
\]

3. Rearrange terms to isolate \( x \):
\[
3x - 6x = 18 + 3
\]
\[
-3x = 21
\]

4. Solve for \( x \):
\[
x = \frac{21}{-3} = -7
\]

5. Verify the solution: Substitute \( x = -7 \) back into the original equation:
\[
\frac{3(-7) - 3}{-7 + 3} = \frac{-21 - 3}{-4} = \frac{-24}{-4} = 6
\]
The solution is correct.

Solution for [d]: \( x = -7 \)

---

#### [e] \( \frac{5x + 2}{x - 10} = 1 \)

1. Eliminate the denominator: Multiply both sides by \( x - 10 \) (assuming \( x \neq 10 \)):
\[
5x + 2 = 1(x - 10)
\]

2. Expand and simplify:
\[
5x + 2 = x - 10
\]

3. Rearrange terms to isolate \( x \):
\[
5x - x = -10 - 2
\]
\[
4x = -12
\]

4. Solve for \( x \):
\[
x = \frac{-12}{4} = -3
\]

5. Verify the solution: Substitute \( x = -3 \) back into the original equation:
\[
\frac{5(-3) + 2}{-3 - 10} = \frac{-15 + 2}{-13} = \frac{-13}{-13} = 1
\]
The solution is correct.

Solution for [e]: \( x = -3 \)

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Q2: Solve the following equations

#### [a] \( \frac{3}{x + 2} = \frac{1}{x + 4} \)

1. Cross-multiply:
\[
3(x + 4) = 1(x + 2)
\]

2. Expand and simplify:
\[
3x + 12 = x + 2
\]

3. Rearrange terms to isolate \( x \):
\[
3x - x = 2 - 12
\]
\[
2x = -10
\]

4. Solve for \( x \):
\[
x = \frac{-10}{2} = -5
\]

5. Verify the solution: Substitute \( x = -5 \) back into the original equation:
\[
\frac{3}{-5 + 2} = \frac{3}{-3} = -1
\]
\[
\frac{1}{-5 + 4} = \frac{1}{-1} = -1
\]
The solution is correct.

Solution for [a]: \( x = -5 \)

---

#### [b] \( \frac{4}{x - 2} = \frac{2}{x - 4} \)

1. Cross-multiply:
\[
4(x - 4) = 2(x - 2)
\]

2. Expand and simplify:
\[
4x - 16 = 2x - 4
\]

3. Rearrange terms to isolate \( x \):
\[
4x - 2x = -4 + 16
\]
\[
2x = 12
\]

4. Solve for \( x \):
\[
x = \frac{12}{2} = 6
\]

5. Verify the solution: Substitute \( x = 6 \) back into the original equation:
\[
\frac{4}{6 - 2} = \frac{4}{4} = 1
\]
\[
\frac{2}{6 - 4} = \frac{2}{2} = 1
\]
The solution is correct.

Solution for [b]: \( x = 6 \)

---

#### [c] \( \frac{1}{x - 3} = \frac{2}{x + 6} \)

1. Cross-multiply:
\[
1(x + 6) = 2(x - 3)
\]

2. Expand and simplify:
\[
x + 6 = 2x - 6
\]

3. Rearrange terms to isolate \( x \):
\[
x - 2x = -6 - 6
\]
\[
-x = -12
\]

4. Solve for \( x \):
\[
x = 12
\]

5. Verify the solution: Substitute \( x = 12 \) back into the original equation:
\[
\frac{1}{12 - 3} = \frac{1}{9}
\]
\[
\frac{2}{12 + 6} = \frac{2}{18} = \frac{1}{9}
\]
The solution is correct.

Solution for [c]: \( x = 12 \)

---

#### [d] \( \frac{2}{x + 1} = \frac{3}{x - 1} \)

1. Cross-multiply:
\[
2(x - 1) = 3(x + 1)
\]

2. Expand and simplify:
\[
2x - 2 = 3x + 3
\]

3. Rearrange terms to isolate \( x \):
\[
2x - 3x = 3 + 2
\]
\[
-x = 5
\]

4. Solve for \( x \):
\[
x = -5
\]

5. Verify the solution: Substitute \( x = -5 \) back into the original equation:
\[
\frac{2}{-5 + 1} = \frac{2}{-4} = -\frac{1}{2}
\]
\[
\frac{3}{-5 - 1} = \frac{3}{-6} = -\frac{1}{2}
\]
The solution is correct.

Solution for [d]: \( x = -5 \)

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#### [e] \( \frac{5}{x + 1} = \frac{4}{x - 4} \)

1. Cross-multiply:
\[
5(x - 4) = 4(x + 1)
\]

2. Expand and simplify:
\[
5x - 20 = 4x + 4
\]

3. Rearrange terms to isolate \( x \):
\[
5x - 4x = 4 + 20
\]
\[
x = 24
\]

4. Verify the solution: Substitute \( x = 24 \) back into the original equation:
\[
\frac{5}{24 + 1} = \frac{5}{25} = \frac{1}{5}
\]
\[
\frac{4}{24 - 4} = \frac{4}{20} = \frac{1}{5}
\]
The solution is correct.

Solution for [e]: \( x = 24 \)

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Final Answers:

\[
\boxed{
\begin{aligned}
&\text{Q1: } [a] \, x = 2, \, [b] \, x = 6, \, [c] \, x = -5, \, [d] \, x = -7, \, [e] \, x = -3 \\
&\text{Q2: } [a] \, x = -5, \, [b] \, x = 6, \, [c] \, x = 12, \, [d] \, x = -5, \, [e] \, x = 24
\end{aligned}
}
\]