Math worksheet for practicing addition and subtraction of rational expressions with like and unlike denominators.

Let's solve each of these rational expression problems step by step, starting with the ones that have like denominators, then moving to those with unlike denominators.

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Section 1: With Like Denominators

When denominators are the same, we can directly combine the numerators.

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#### Problem ①
$$
\frac{4x + 4}{x^2 - 2x - 8} - \frac{3x + 2}{x^2 - 2x - 8}
$$

Since denominators are the same:
$$
= \frac{(4x + 4) - (3x + 2)}{x^2 - 2x - 8}
= \frac{4x + 4 - 3x - 2}{x^2 - 2x - 8}
= \frac{x + 2}{x^2 - 2x - 8}
$$

Now factor the denominator:
$$
x^2 - 2x - 8 = (x - 4)(x + 2)
$$

So:
$$
\frac{x + 2}{(x - 4)(x + 2)} = \frac{1}{x - 4}, \quad \text{for } x \neq -2
$$

✔ Answer: $\boxed{\frac{1}{x - 4}}$

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#### Problem ②
$$
\frac{2x^2 + 3}{x^2 - 6x + 5} - \frac{x^2 - 5x + 9}{x^2 - 6x + 5}
$$

Same denominator:
$$
= \frac{(2x^2 + 3) - (x^2 - 5x + 9)}{x^2 - 6x + 5}
= \frac{2x^2 + 3 - x^2 + 5x - 9}{x^2 - 6x + 5}
= \frac{x^2 + 5x - 6}{x^2 - 6x + 5}
$$

Factor numerator and denominator:

Numerator: $x^2 + 5x - 6 = (x + 6)(x - 1)$
Denominator: $x^2 - 6x + 5 = (x - 5)(x - 1)$

So:
$$
\frac{(x + 6)(x - 1)}{(x - 5)(x - 1)} = \frac{x + 6}{x - 5}, \quad \text{for } x \neq 1
$$

✔ Answer: $\boxed{\frac{x + 6}{x - 5}}$

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#### Problem ③
$$
\frac{3v - 4}{2v^2 + 8v + 6} - \frac{2v - 7}{2v^2 + 8v + 6}
$$

Same denominator:
$$
= \frac{(3v - 4) - (2v - 7)}{2v^2 + 8v + 6}
= \frac{3v - 4 - 2v + 7}{2v^2 + 8v + 6}
= \frac{v + 3}{2v^2 + 8v + 6}
$$

Factor denominator:
$$
2v^2 + 8v + 6 = 2(v^2 + 4v + 3) = 2(v + 1)(v + 3)
$$

So:
$$
\frac{v + 3}{2(v + 1)(v + 3)} = \frac{1}{2(v + 1)}, \quad \text{for } v \neq -3
$$

✔ Answer: $\boxed{\frac{1}{2(v + 1)}}$

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#### Problem ④
$$
\frac{3k - 2}{3k^2 - 19k + 6} + \frac{1}{3k^2 - 19k + 6}
$$

Same denominator:
$$
= \frac{(3k - 2) + 1}{3k^2 - 19k + 6}
= \frac{3k - 1}{3k^2 - 19k + 6}
$$

Factor denominator:
We need two numbers that multiply to $3 \cdot 6 = 18$ and add to $-19$.
Try: $-18$ and $-1$: $-18 \cdot -1 = 18$, $-18 -1 = -19$ → yes!

So:
$$
3k^2 - 19k + 6 = 3k^2 - 18k - k + 6 = 3k(k - 6) -1(k - 6) = (3k - 1)(k - 6)
$$

So:
$$
\frac{3k - 1}{(3k - 1)(k - 6)} = \frac{1}{k - 6}, \quad \text{for } k \neq \frac{1}{3}
$$

✔ Answer: $\boxed{\frac{1}{k - 6}}$

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Section 2: With Unlike Denominators

We need to find a common denominator, simplify, and combine.

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#### Problem ⑤
$$
\frac{4xy}{x^2 - y^2} + \frac{x - y}{x + y}
$$

Note: $x^2 - y^2 = (x - y)(x + y)$

So rewrite first fraction:
$$
\frac{4xy}{(x - y)(x + y)} + \frac{x - y}{x + y}
$$

Common denominator is $(x - y)(x + y)$

Rewrite second term:
$$
\frac{x - y}{x + y} = \frac{(x - y)(x - y)}{(x + y)(x - y)} = \frac{(x - y)^2}{(x - y)(x + y)}
$$

Now add:
$$
\frac{4xy + (x - y)^2}{(x - y)(x + y)}
$$

Expand $(x - y)^2 = x^2 - 2xy + y^2$

So numerator:
$$
4xy + x^2 - 2xy + y^2 = x^2 + 2xy + y^2 = (x + y)^2
$$

So:
$$
\frac{(x + y)^2}{(x - y)(x + y)} = \frac{x + y}{x - y}, \quad \text{for } x \neq y
$$

✔ Answer: $\boxed{\frac{x + y}{x - y}}$

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#### Problem ⑥
$$
\frac{2}{a^2 - 5a + 4} + \frac{-2}{a^2 - 4}
$$

Factor denominators:

- $a^2 - 5a + 4 = (a - 1)(a - 4)$
- $a^2 - 4 = (a - 2)(a + 2)$

So LCD is $(a - 1)(a - 4)(a - 2)(a + 2)$ — but let’s see if we can keep it simpler or just proceed.

Write both fractions:

$$
\frac{2}{(a - 1)(a - 4)} + \frac{-2}{(a - 2)(a + 2)}
$$

Find LCD: $(a - 1)(a - 4)(a - 2)(a + 2)$

Now rewrite each:

First term:
$$
\frac{2(a - 2)(a + 2)}{(a - 1)(a - 4)(a - 2)(a + 2)}
= \frac{2(a^2 - 4)}{(a - 1)(a - 4)(a - 2)(a + 2)}
$$

Second term:
$$
\frac{-2(a - 1)(a - 4)}{(a - 1)(a - 4)(a - 2)(a + 2)}
= \frac{-2(a - 1)(a - 4)}{(a - 1)(a - 4)(a - 2)(a + 2)}
$$

Now combine:
$$
\frac{2(a^2 - 4) - 2(a - 1)(a - 4)}{(a - 1)(a - 4)(a - 2)(a + 2)}
$$

Compute numerator:

1. $2(a^2 - 4) = 2a^2 - 8$
2. $(a - 1)(a - 4) = a^2 - 5a + 4$, so $-2(a^2 - 5a + 4) = -2a^2 + 10a - 8$

Add:
$$
(2a^2 - 8) + (-2a^2 + 10a - 8) = 10a - 16
$$

So:
$$
\frac{10a - 16}{(a - 1)(a - 4)(a - 2)(a + 2)}
= \frac{2(5a - 8)}{(a - 1)(a - 4)(a - 2)(a + 2)}
$$

No further simplification possible.

✔ Answer: $\boxed{\frac{2(5a - 8)}{(a - 1)(a - 4)(a - 2)(a + 2)}}$

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#### Problem ⑦
$$
2 + \frac{x - 3}{x + 1}
$$

Write 2 as a fraction with denominator $x + 1$:
$$
= \frac{2(x + 1)}{x + 1} + \frac{x - 3}{x + 1}
= \frac{2x + 2 + x - 3}{x + 1}
= \frac{3x - 1}{x + 1}
$$

✔ Answer: $\boxed{\frac{3x - 1}{x + 1}}$

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#### Problem ⑧
$$
\frac{4x}{x^2 - 2x - 3} - \frac{3}{x^2 - 5x + 6}
$$

Factor denominators:

- $x^2 - 2x - 3 = (x - 3)(x + 1)$
- $x^2 - 5x + 6 = (x - 2)(x - 3)$

LCD: $(x - 3)(x + 1)(x - 2)$

Rewrite each fraction:

First:
$$
\frac{4x}{(x - 3)(x + 1)} = \frac{4x(x - 2)}{(x - 3)(x + 1)(x - 2)}
$$

Second:
$$
\frac{3}{(x - 2)(x - 3)} = \frac{3(x + 1)}{(x - 2)(x - 3)(x + 1)}
$$

Now subtract:
$$
\frac{4x(x - 2) - 3(x + 1)}{(x - 3)(x + 1)(x - 2)}
$$

Compute numerator:

- $4x(x - 2) = 4x^2 - 8x$
- $3(x + 1) = 3x + 3$
- So: $4x^2 - 8x - (3x + 3) = 4x^2 - 11x - 3$

So:
$$
\frac{4x^2 - 11x - 3}{(x - 3)(x + 1)(x - 2)}
$$

Check if numerator factors:
Try quadratic formula:
$$
x = \frac{11 \pm \sqrt{(-11)^2 - 4(4)(-3)}}{2(4)} = \frac{11 \pm \sqrt{121 + 48}}{8} = \frac{11 \pm \sqrt{169}}{8} = \frac{11 \pm 13}{8}
$$

So roots: $x = \frac{24}{8} = 3$, $x = \frac{-2}{8} = -\frac{1}{4}$

Thus: $4x^2 - 11x - 3 = (x - 3)(4x + 1)$

Wait: Let's check: $(x - 3)(4x + 1) = 4x^2 + x - 12x - 3 = 4x^2 - 11x - 3$ ✔

So numerator: $(x - 3)(4x + 1)$

Denominator: $(x - 3)(x + 1)(x - 2)$

Cancel $x - 3$:
$$
\frac{4x + 1}{(x + 1)(x - 2)}, \quad \text{for } x \neq 3
$$

✔ Answer: $\boxed{\frac{4x + 1}{(x + 1)(x - 2)}}$

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✔ Final Answers Summary:

1. $\boxed{\frac{1}{x - 4}}$
2. $\boxed{\frac{x + 6}{x - 5}}$
3. $\boxed{\frac{1}{2(v + 1)}}$
4. $\boxed{\frac{1}{k - 6}}$
5. $\boxed{\frac{x + y}{x - y}}$
6. $\boxed{\frac{2(5a - 8)}{(a - 1)(a - 4)(a - 2)(a + 2)}}$
7. $\boxed{\frac{3x - 1}{x + 1}}$
8. $\boxed{\frac{4x + 1}{(x + 1)(x - 2)}}$

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