Solving Equations with fractions. - Free Printable

Let's solve each equation step-by-step. We'll go through Q1 and Q2 systematically, using algebraic methods to solve equations involving fractions.

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Q1: Solve the following equations

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

Step 1: Multiply both sides by $x - 1$ (the denominator) to eliminate the fraction:
$$
2x + 3 = 7(x - 1)
$$

Step 2: Expand the right-hand side:
$$
2x + 3 = 7x - 7
$$

Step 3: Bring all terms to one side:
$$
2x + 3 - 7x + 7 = 0 \Rightarrow -5x + 10 = 0
$$

Step 4: Solve for $x$:
$$
-5x = -10 \Rightarrow x = 2
$$

✔ Check: Plug $x = 2$ into original equation:
$$
\frac{2(2) + 3}{2 - 1} = \frac{4 + 3}{1} = 7 \quad \text{✓}
$$

> ✔ Answer: $x = 2$

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

Step 1: Multiply both sides by $x + 5$:
$$
4x - 2 = 2(x + 5)
$$

Step 2: Expand:
$$
4x - 2 = 2x + 10
$$

Step 3: Solve:
$$
4x - 2x = 10 + 2 \Rightarrow 2x = 12 \Rightarrow x = 6
$$

✔ Check: $\frac{4(6) - 2}{6 + 5} = \frac{24 - 2}{11} = \frac{22}{11} = 2$ ✓

> ✔ Answer: $x = 6$

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

Step 1: Multiply both sides by $x - 2$:
$$
5x - 3 = 4(x - 2)
$$

Step 2: Expand:
$$
5x - 3 = 4x - 8
$$

Step 3: Solve:
$$
5x - 4x = -8 + 3 \Rightarrow x = -5
$$

✔ Check: $\frac{5(-5) - 3}{-5 - 2} = \frac{-25 - 3}{-7} = \frac{-28}{-7} = 4$ ✓

> ✔ Answer: $x = -5$

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

Step 1: Multiply both sides by $x + 3$:
$$
3x - 3 = 6(x + 3)
$$

Step 2: Expand:
$$
3x - 3 = 6x + 18
$$

Step 3: Solve:
$$
3x - 6x = 18 + 3 \Rightarrow -3x = 21 \Rightarrow x = -7
$$

✔ Check: $\frac{3(-7) - 3}{-7 + 3} = \frac{-21 - 3}{-4} = \frac{-24}{-4} = 6$ ✓

> ✔ Answer: $x = -7$

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

Step 1: Multiply both sides by $x - 10$:
$$
5x + 2 = 1(x - 10)
$$

Step 2: Simplify:
$$
5x + 2 = x - 10
$$

Step 3: Solve:
$$
5x - x = -10 - 2 \Rightarrow 4x = -12 \Rightarrow x = -3
$$

✔ Check: $\frac{5(-3) + 2}{-3 - 10} = \frac{-15 + 2}{-13} = \frac{-13}{-13} = 1$ ✓

> ✔ Answer: $x = -3$

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Q2: Solve the following equations

These are equations with two fractions, so we’ll cross-multiply.

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

Step 1: Cross-multiply:
$$
3(x + 4) = 1(x + 2)
$$

Step 2: Expand:
$$
3x + 12 = x + 2
$$

Step 3: Solve:
$$
3x - x = 2 - 12 \Rightarrow 2x = -10 \Rightarrow x = -5
$$

✔ Check:
Left: $\frac{3}{-5 + 2} = \frac{3}{-3} = -1$
Right: $\frac{1}{-5 + 4} = \frac{1}{-1} = -1$ ✓

> ✔ Answer: $x = -5$

---

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

Step 1: Cross-multiply:
$$
4(x - 4) = 2(x - 2)
$$

Step 2: Expand:
$$
4x - 16 = 2x - 4
$$

Step 3: Solve:
$$
4x - 2x = -4 + 16 \Rightarrow 2x = 12 \Rightarrow x = 6
$$

✔ Check:
Left: $\frac{4}{6 - 2} = \frac{4}{4} = 1$
Right: $\frac{2}{6 - 4} = \frac{2}{2} = 1$ ✓

> ✔ Answer: $x = 6$

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

Step 1: Cross-multiply:
$$
1(x + 6) = 2(x - 3)
$$

Step 2: Expand:
$$
x + 6 = 2x - 6
$$

Step 3: Solve:
$$
6 + 6 = 2x - x \Rightarrow 12 = x \Rightarrow x = 12
$$

✔ Check:
Left: $\frac{1}{12 - 3} = \frac{1}{9}$
Right: $\frac{2}{12 + 6} = \frac{2}{18} = \frac{1}{9}$ ✓

> ✔ Answer: $x = 12$

---

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

Step 1: Cross-multiply:
$$
2(x - 1) = 3(x + 1)
$$

Step 2: Expand:
$$
2x - 2 = 3x + 3
$$

Step 3: Solve:
$$
-2 - 3 = 3x - 2x \Rightarrow -5 = x \Rightarrow x = -5
$$

✔ Check:
Left: $\frac{2}{-5 + 1} = \frac{2}{-4} = -\frac{1}{2}$
Right: $\frac{3}{-5 - 1} = \frac{3}{-6} = -\frac{1}{2}$ ✓

> ✔ Answer: $x = -5$

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

Step 1: Cross-multiply:
$$
5(x - 4) = 4(x + 1)
$$

Step 2: Expand:
$$
5x - 20 = 4x + 4
$$

Step 3: Solve:
$$
5x - 4x = 4 + 20 \Rightarrow x = 24
$$

✔ Check:
Left: $\frac{5}{24 + 1} = \frac{5}{25} = \frac{1}{5}$
Right: $\frac{4}{24 - 4} = \frac{4}{20} = \frac{1}{5}$ ✓

> ✔ Answer: $x = 24$

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✔ Final Answers Summary:

#### Q1:
- [a] $x = 2$
- [b] $x = 6$
- [c] $x = -5$
- [d] $x = -7$
- [e] $x = -3$

#### Q2:
- [a] $x = -5$
- [b] $x = 6$
- [c] $x = 12$
- [d] $x = -5$
- [e] $x = 24$

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🔍 Key Tips:

- Always check for restrictions: Denominators cannot be zero.
- For example, in Q1[a], $x \neq 1$, and our solution $x = 2$ is valid.
- When solving equations with two fractions, cross-multiplication is often the quickest method.
- Always verify your solution by plugging back into the original equation.

Let me know if you'd like these solved graphically or with additional explanation!