Algebraic fractions practice problems with step-by-step solutions.

You’ve uploaded an image titled “Algebraic Fractions: 1”, which presents a set of problems (A through V) involving adding or subtracting algebraic fractions. Each problem shows the original expression, then the step of rewriting with a common denominator, and finally leaves a blank for the simplified result.

Let’s solve all of them step by step and explain the process.

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🔑 General Strategy:

To add or subtract fractions:
1. Find the Least Common Denominator (LCD).
2. Rewrite each fraction with the LCD as the denominator.
3. Combine numerators.
4. Simplify if possible.

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## ✔ A) `x/2 + x/3`
→ `x(3)/6 + x(2)/6` = `(3x + 2x)/6` = `5x/6`

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## ✔ B) `x/4 + x/5`
LCD = 20
→ `x(5)/20 + x(4)/20` = `(5x + 4x)/20` = `9x/20`

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## ✔ C) `x/3 - x/5`
LCD = 15
→ `x(5)/15 - x(3)/15` = `(5x - 3x)/15` = `2x/15`

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## ✔ D) `1/2 + x/3`
LCD = 6
→ `1(3)/6 + x(2)/6` = `(3 + 2x)/6` → Already shown; final answer is `(2x + 3)/6`

*(Note: Order doesn’t matter — `2x + 3` is same as `3 + 2x`)*

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## ✔ E) `1/4 + x/5`
LCD = 20
→ `1(5)/20 + x(4)/20` = `(5 + 4x)/20` → `(4x + 5)/20`

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## ✔ F) `1/3 - x/2`
LCD = 6
→ `1(2)/6 - x(3)/6` = `(2 - 3x)/6` → `(2 - 3x)/6`

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## ✔ G) `1/2 + 2x/3`
LCD = 6
→ `1(3)/6 + 2x(2)/6` = `(3 + 4x)/6` → `(4x + 3)/6`

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## ✔ H) `1/4 + 3x/5`
LCD = 20
→ `1(5)/20 + 3x(4)/20` = `(5 + 12x)/20` → `(12x + 5)/20`

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## ✔ I) `1/4 - 2x/3`
LCD = 12
→ `1(3)/12 - 2x(4)/12` = `(3 - 8x)/12` → `(3 - 8x)/12`

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## ✔ J) `1/2 + 4/x`
LCD = `2x`
→ `1(x)/(2x) + 4(2)/(2x)` = `(x + 8)/(2x)` → `(x + 8)/(2x)`

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## ✔ M) `1/2 + 4/(3x)`
LCD = `6x`
→ `1(3x)/(6x) + 4(2)/(6x)` = `(3x + 8)/(6x)` → `(3x + 8)/(6x)`

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## ✔ N) `1/4 + 3/(5x)`
LCD = `20x`
→ `1(5x)/(20x) + 3(4)/(20x)` = `(5x + 12)/(20x)` → `(5x + 12)/(20x)`

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## ✔ O) `1/5 - 3/(4x)`
LCD = `20x`
→ `1(4x)/(20x) - 3(5)/(20x)` = `(4x - 15)/(20x)` → `(4x - 15)/(20x)`

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## ✔ P) `x/2 + 1/x`
LCD = `2x`
→ `x(x)/(2x) + 1(2)/(2x)` = `(x² + 2)/(2x)` → `(x² + 2)/(2x)`

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## ✔ Q) `x/4 + 3/x`
LCD = `4x`
→ `x(x)/(4x) + 3(4)/(4x)` = `(x² + 12)/(4x)` → `(x² + 12)/(4x)`

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## ✔ R) `x/5 - 2/x`
LCD = `5x`
→ `x(x)/(5x) - 2(5)/(5x)` = `(x² - 10)/(5x)` → `(x² - 10)/(5x)`

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## ✔ S) `x/2 + 1/(3x)`
LCD = `6x`
→ `x(3x)/(6x) + 1(2)/(6x)` = `(3x² + 2)/(6x)` → `(3x² + 2)/(6x)`

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## ✔ T) `x/3 + 3/(4x)`
LCD = `12x`
→ `x(4x)/(12x) + 3(3)/(12x)` = `(4x² + 9)/(12x)` → `(4x² + 9)/(12x)`

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## ✔ U) `x/3 - 5/(2x)`
LCD = `6x`
→ `x(2x)/(6x) - 5(3)/(6x)` = `(2x² - 15)/(6x)` → `(2x² - 15)/(6x)`

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## ✔ V) `1/(3x) + 2x/5`
LCD = `15x`
→ `1(5)/(15x) + 2x(3x)/(15x)` = `(5 + 6x²)/(15x)` → `(6x² + 5)/(15x)`

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## 📝 Final Answers Summary:

| Problem | Answer |
|---------|------------------------|
| A | `5x/6` |
| B | `9x/20` |
| C | `2x/15` |
| D | `(2x + 3)/6` |
| E | `(4x + 5)/20` |
| F | `(2 - 3x)/6` |
| G | `(4x + 3)/6` |
| H | `(12x + 5)/20` |
| I | `(3 - 8x)/12` |
| J | `(x + 8)/(2x)` |
| M | `(3x + 8)/(6x)` |
| N | `(5x + 12)/(20x)` |
| O | `(4x - 15)/(20x)` |
| P | `(x² + 2)/(2x)` |
| Q | `(x² + 12)/(4x)` |
| R | `(x² - 10)/(5x)` |
| S | `(3x² + 2)/(6x)` |
| T | `(4x² + 9)/(12x)` |
| U | `(2x² - 15)/(6x)` |
| V | `(6x² + 5)/(15x)` |

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✔ All problems are solved! The key takeaway is to always find the LCD, rewrite each term, combine numerators, and simplify when possible. Remember to keep variables in order (usually descending powers) for neatness.

Let me know if you’d like to see any step explained in more detail!