9th Grade Algebra I Pace Set (High School) from Accelerated ... - Free Printable

You've uploaded a math worksheet that teaches how to add and subtract rational expressions by finding the Least Common Denominator (LCD). The example at the top shows:

> \(\frac{2a}{3b} - \frac{4a}{5b} + \frac{5a}{2b} = ?\)

And it correctly finds the LCD as \(30b\) (since LCM of 3, 5, and 2 is 30, and all denominators have a \(b\)), then rewrites each fraction with denominator \(30b\), combines numerators, and simplifies to \(\frac{71a}{30b}\).

Now, let’s solve the 5 problems below using the same method.

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Problem 1: \(\frac{1}{2} - \frac{1}{3} + \frac{3}{4}\)

Step 1: Find LCD

Denominators: 2, 3, 4
Prime factors:
- 2 → 2
- 3 → 3
- 4 → 2²

LCD = highest powers: \(2^2 \cdot 3 = 12\)

Step 2: Rewrite each fraction with denominator 12

- \(\frac{1}{2} = \frac{6}{12}\)
- \(\frac{1}{3} = \frac{4}{12}\)
- \(\frac{3}{4} = \frac{9}{12}\)

Step 3: Combine

\[
\frac{6}{12} - \frac{4}{12} + \frac{9}{12} = \frac{6 - 4 + 9}{12} = \frac{11}{12}
\]

✔ Answer: \(\boxed{\frac{11}{12}}\)

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Problem 2: \(\frac{2x}{3} + \frac{5x}{6} - \frac{x}{2}\)

Step 1: Find LCD

Denominators: 3, 6, 2
LCM of 3, 6, 2 is 6

Step 2: Rewrite each fraction with denominator 6

- \(\frac{2x}{3} = \frac{4x}{6}\)
- \(\frac{5x}{6} = \frac{5x}{6}\)
- \(\frac{x}{2} = \frac{3x}{6}\)

Step 3: Combine

\[
\frac{4x}{6} + \frac{5x}{6} - \frac{3x}{6} = \frac{4x + 5x - 3x}{6} = \frac{6x}{6} = x
\]

✔ Answer: \(\boxed{x}\)

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Problem 3: \(\frac{1}{a} + \frac{1}{b}\)

Step 1: Find LCD

Since \(a\) and \(b\) are different variables (assumed to be distinct and non-zero), LCD = \(ab\)

Step 2: Rewrite each fraction

- \(\frac{1}{a} = \frac{b}{ab}\)
- \(\frac{1}{b} = \frac{a}{ab}\)

Step 3: Combine

\[
\frac{b}{ab} + \frac{a}{ab} = \frac{a + b}{ab}
\]

✔ Answer: \(\boxed{\frac{a + b}{ab}}\)

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Problem 4: \(\frac{7x}{4a} - \frac{3x}{5a}\)

Step 1: Find LCD

Denominators: \(4a\) and \(5a\) → LCD = LCM(4,5) × a = \(20a\)

Step 2: Rewrite each fraction

- \(\frac{7x}{4a} = \frac{7x \cdot 5}{4a \cdot 5} = \frac{35x}{20a}\)
- \(\frac{3x}{5a} = \frac{3x \cdot 4}{5a \cdot 4} = \frac{12x}{20a}\)

Step 3: Subtract

\[
\frac{35x}{20a} - \frac{12x}{20a} = \frac{23x}{20a}
\]

✔ Answer: \(\boxed{\frac{23x}{20a}}\)

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Problem 5: \(\frac{2}{b^2} + \frac{3}{b}\)

Step 1: Find LCD

Denominators: \(b^2\) and \(b\) → LCD = \(b^2\)

Step 2: Rewrite each fraction

- \(\frac{2}{b^2} = \frac{2}{b^2}\)
- \(\frac{3}{b} = \frac{3b}{b^2}\)

Step 3: Add

\[
\frac{2}{b^2} + \frac{3b}{b^2} = \frac{2 + 3b}{b^2}
\]

✔ Answer: \(\boxed{\frac{3b + 2}{b^2}}\) (or \(\frac{2 + 3b}{b^2}\), same thing)

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## ✔ Final Answers:

1. \(\boxed{\frac{11}{12}}\)

2. \(\boxed{x}\)

3. \(\boxed{\frac{a + b}{ab}}\)

4. \(\boxed{\frac{23x}{20a}}\)

5. \(\boxed{\frac{3b + 2}{b^2}}\)

Let me know if you’d like to see any step explained further!