Adding and subtracting rational expressions with unlike denominators - Free Printable

Let's solve each of these rational expression problems step by step. We'll simplify each expression by finding a common denominator and combining the terms.

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Problem 27:

$$
\frac{9}{2y + 3} - \frac{5}{4y}
$$

Step 1: Identify denominators
- Denominators: $2y + 3$ and $4y$

These are not factorable in a way that shares common factors, so the least common denominator (LCD) is:
$$
(2y + 3)(4y)
$$

Step 2: Rewrite each fraction with the LCD

$$
\frac{9}{2y + 3} = \frac{9 \cdot 4y}{(2y + 3)(4y)} = \frac{36y}{(2y + 3)(4y)}
$$

$$
\frac{5}{4y} = \frac{5 \cdot (2y + 3)}{(4y)(2y + 3)} = \frac{10y + 15}{(2y + 3)(4y)}
$$

Step 3: Subtract

$$
\frac{36y - (10y + 15)}{(2y + 3)(4y)} = \frac{36y - 10y - 15}{(2y + 3)(4y)} = \frac{26y - 15}{(2y + 3)(4y)}
$$

Final Answer:
$$
\boxed{\frac{26y - 15}{4y(2y + 3)}}
$$

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Problem 28:

$$
\frac{4}{5x + 7} + \frac{7}{6x}
$$

Step 1: Denominators:
- $5x + 7$ and $6x$ — no common factors, so LCD is $(5x + 7)(6x)$

Step 2: Rewrite fractions

$$
\frac{4}{5x + 7} = \frac{4 \cdot 6x}{(5x + 7)(6x)} = \frac{24x}{(5x + 7)(6x)}
$$

$$
\frac{7}{6x} = \frac{7 \cdot (5x + 7)}{(6x)(5x + 7)} = \frac{35x + 49}{(5x + 7)(6x)}
$$

Step 3: Add

$$
\frac{24x + 35x + 49}{(5x + 7)(6x)} = \frac{59x + 49}{(5x + 7)(6x)}
$$

Final Answer:
$$
\boxed{\frac{59x + 49}{6x(5x + 7)}}
$$

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Problem 29:

$$
\frac{1}{3x^2 - 8x} + \frac{6}{8 - 3x}
$$

Step 1: Factor denominators

- $3x^2 - 8x = x(3x - 8)$
- $8 - 3x = -(3x - 8)$

So rewrite the second term:

$$
\frac{6}{8 - 3x} = \frac{6}{-(3x - 8)} = -\frac{6}{3x - 8}
$$

Now the expression becomes:

$$
\frac{1}{x(3x - 8)} - \frac{6}{3x - 8}
$$

Step 2: LCD is $x(3x - 8)$

Rewrite second term:

$$
-\frac{6}{3x - 8} = -\frac{6x}{x(3x - 8)}
$$

Now combine:

$$
\frac{1 - 6x}{x(3x - 8)}
$$

Final Answer:
$$
\boxed{\frac{1 - 6x}{x(3x - 8)}}
$$

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Problem 30:

$$
\frac{x + 4}{3x^2 - 4x} - \frac{6}{4 - 3x}
$$

Step 1: Factor denominators

- $3x^2 - 4x = x(3x - 4)$
- $4 - 3x = -(3x - 4)$

So:
$$
\frac{6}{4 - 3x} = \frac{6}{-(3x - 4)} = -\frac{6}{3x - 4}
$$

Now the expression becomes:

$$
\frac{x + 4}{x(3x - 4)} - \left(-\frac{6}{3x - 4}\right) = \frac{x + 4}{x(3x - 4)} + \frac{6}{3x - 4}
$$

Step 2: LCD is $x(3x - 4)$

Rewrite second term:

$$
\frac{6}{3x - 4} = \frac{6x}{x(3x - 4)}
$$

Now add:

$$
\frac{x + 4 + 6x}{x(3x - 4)} = \frac{7x + 4}{x(3x - 4)}
$$

Final Answer:
$$
\boxed{\frac{7x + 4}{x(3x - 4)}}
$$

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✔ Final Answers:

27. $\boxed{\frac{26y - 15}{4y(2y + 3)}}$
28. $\boxed{\frac{59x + 49}{6x(5x + 7)}}$
29. $\boxed{\frac{1 - 6x}{x(3x - 8)}}$
30. $\boxed{\frac{7x + 4}{x(3x - 4)}}$