Solving Equations with fractions. - Free Printable

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 by multiplying both sides by \( x - 1 \):
\[
\frac{2x + 3}{x - 1} \cdot (x - 1) = 7 \cdot (x - 1)
\]
\[
2x + 3 = 7(x - 1)
\]

2. Expand the right-hand side:
\[
2x + 3 = 7x - 7
\]

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

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

5. Check for extraneous solutions: Substitute \( x = 2 \) back into the original equation to ensure it does not make the denominator zero.
\[
\text{Denominator: } x - 1 = 2 - 1 = 1 \neq 0
\]
The solution is valid.

Solution: \( x = 2 \)

---

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

1. Eliminate the denominator by multiplying both sides by \( x + 5 \):
\[
\frac{4x - 2}{x + 5} \cdot (x + 5) = 2 \cdot (x + 5)
\]
\[
4x - 2 = 2(x + 5)
\]

2. Expand the right-hand side:
\[
4x - 2 = 2x + 10
\]

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

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

5. Check for extraneous solutions: Substitute \( x = 6 \) back into the original equation to ensure it does not make the denominator zero.
\[
\text{Denominator: } x + 5 = 6 + 5 = 11 \neq 0
\]
The solution is valid.

Solution: \( x = 6 \)

---

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

1. Eliminate the denominator by multiplying both sides by \( x - 2 \):
\[
\frac{5x - 3}{x - 2} \cdot (x - 2) = 4 \cdot (x - 2)
\]
\[
5x - 3 = 4(x - 2)
\]

2. Expand the right-hand side:
\[
5x - 3 = 4x - 8
\]

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

4. Check for extraneous solutions: Substitute \( x = -5 \) back into the original equation to ensure it does not make the denominator zero.
\[
\text{Denominator: } x - 2 = -5 - 2 = -7 \neq 0
\]
The solution is valid.

Solution: \( x = -5 \)

---

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

1. Eliminate the denominator by multiplying both sides by \( x + 3 \):
\[
\frac{3x - 3}{x + 3} \cdot (x + 3) = 6 \cdot (x + 3)
\]
\[
3x - 3 = 6(x + 3)
\]

2. Expand the right-hand side:
\[
3x - 3 = 6x + 18
\]

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

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

5. Check for extraneous solutions: Substitute \( x = -7 \) back into the original equation to ensure it does not make the denominator zero.
\[
\text{Denominator: } x + 3 = -7 + 3 = -4 \neq 0
\]
The solution is valid.

Solution: \( x = -7 \)

---

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

1. Eliminate the denominator by multiplying both sides by \( x - 10 \):
\[
\frac{5x + 2}{x - 10} \cdot (x - 10) = 1 \cdot (x - 10)
\]
\[
5x + 2 = x - 10
\]

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

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

4. Check for extraneous solutions: Substitute \( x = -3 \) back into the original equation to ensure it does not make the denominator zero.
\[
\text{Denominator: } x - 10 = -3 - 10 = -13 \neq 0
\]
The solution is valid.

Solution: \( x = -3 \)

---

Q2: Solve the following equations

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

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

2. Expand both sides:
\[
3x + 12 = x + 2
\]

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

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

5. Check for extraneous solutions: Substitute \( x = -5 \) back into the original equation to ensure it does not make any denominator zero.
\[
\text{Denominators: } x + 2 = -5 + 2 = -3 \neq 0 \quad \text{and} \quad x + 4 = -5 + 4 = -1 \neq 0
\]
The solution is valid.

Solution: \( x = -5 \)

---

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

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

2. Expand both sides:
\[
4x - 16 = 2x - 4
\]

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

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

5. Check for extraneous solutions: Substitute \( x = 6 \) back into the original equation to ensure it does not make any denominator zero.
\[
\text{Denominators: } x - 2 = 6 - 2 = 4 \neq 0 \quad \text{and} \quad x - 4 = 6 - 4 = 2 \neq 0
\]
The solution is valid.

Solution: \( x = 6 \)

---

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

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

2. Expand both sides:
\[
x + 6 = 2x - 6
\]

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

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

5. Check for extraneous solutions: Substitute \( x = 12 \) back into the original equation to ensure it does not make any denominator zero.
\[
\text{Denominators: } x - 3 = 12 - 3 = 9 \neq 0 \quad \text{and} \quad x + 6 = 12 + 6 = 18 \neq 0
\]
The solution is valid.

Solution: \( x = 12 \)

---

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

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

2. Expand both sides:
\[
2x - 2 = 3x + 3
\]

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

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

5. Check for extraneous solutions: Substitute \( x = -5 \) back into the original equation to ensure it does not make any denominator zero.
\[
\text{Denominators: } x + 1 = -5 + 1 = -4 \neq 0 \quad \text{and} \quad x - 1 = -5 - 1 = -6 \neq 0
\]
The solution is valid.

Solution: \( x = -5 \)

---

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

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

2. Expand both sides:
\[
5x - 20 = 4x + 4
\]

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

4. Check for extraneous solutions: Substitute \( x = 24 \) back into the original equation to ensure it does not make any denominator zero.
\[
\text{Denominators: } x + 1 = 24 + 1 = 25 \neq 0 \quad \text{and} \quad x - 4 = 24 - 4 = 20 \neq 0
\]
The solution is valid.

Solution: \( x = 24 \)

---

Final Answers:

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