Rational Expression Worksheet: Adding and Subtracting Algebraic Fractions
Rational expression worksheet for adding and subtracting fractions with variables, featuring eight problems involving algebraic expressions.
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Show Answer Key & Explanations
Step-by-step solution for: Adding and Subtracting Rational Expressions Worksheets - Math Monks
▼
Show Answer Key & Explanations
Step-by-step solution for: Adding and Subtracting Rational Expressions Worksheets - Math Monks
To solve the given rational expressions, we need to follow these steps:
1. Identify common denominators when adding or subtracting.
2. Combine numerators after ensuring the denominators are the same.
3. Simplify the resulting expression if possible.
Let's solve each problem step by step.
---
$$
\frac{7}{y-5} - \frac{4}{y-5}
$$
- The denominators are the same: \( y-5 \).
- Combine the numerators:
$$
\frac{7 - 4}{y-5} = \frac{3}{y-5}
$$
Solution:
$$
\boxed{\frac{3}{y-5}}
$$
---
$$
\frac{4}{a+1} + \frac{2}{a+1}
$$
- The denominators are the same: \( a+1 \).
- Combine the numerators:
$$
\frac{4 + 2}{a+1} = \frac{6}{a+1}
$$
Solution:
$$
\boxed{\frac{6}{a+1}}
$$
---
$$
\frac{5}{3x^2} - \frac{1}{3x^2}
$$
- The denominators are the same: \( 3x^2 \).
- Combine the numerators:
$$
\frac{5 - 1}{3x^2} = \frac{4}{3x^2}
$$
Solution:
$$
\boxed{\frac{4}{3x^2}}
$$
---
$$
\frac{6}{x-1} - \frac{5x}{4}
$$
- The denominators are different: \( x-1 \) and \( 4 \).
- Find the least common denominator (LCD): \( 4(x-1) \).
- Rewrite each fraction with the LCD:
$$
\frac{6}{x-1} = \frac{6 \cdot 4}{(x-1) \cdot 4} = \frac{24}{4(x-1)}
$$
$$
\frac{5x}{4} = \frac{5x \cdot (x-1)}{4 \cdot (x-1)} = \frac{5x(x-1)}{4(x-1)}
$$
- Combine the fractions:
$$
\frac{24}{4(x-1)} - \frac{5x(x-1)}{4(x-1)} = \frac{24 - 5x(x-1)}{4(x-1)}
$$
- Simplify the numerator:
$$
24 - 5x(x-1) = 24 - 5x^2 + 5x
$$
- Final expression:
$$
\frac{-5x^2 + 5x + 24}{4(x-1)}
$$
Solution:
$$
\boxed{\frac{-5x^2 + 5x + 24}{4(x-1)}}
$$
---
$$
\frac{5}{x+5} + \frac{4x}{2x+6}
$$
- Simplify the second fraction:
$$
\frac{4x}{2x+6} = \frac{4x}{2(x+3)} = \frac{2x}{x+3}
$$
- The denominators are now \( x+5 \) and \( x+3 \).
- Find the least common denominator (LCD): \( (x+5)(x+3) \).
- Rewrite each fraction with the LCD:
$$
\frac{5}{x+5} = \frac{5(x+3)}{(x+5)(x+3)} = \frac{5x + 15}{(x+5)(x+3)}
$$
$$
\frac{2x}{x+3} = \frac{2x(x+5)}{(x+3)(x+5)} = \frac{2x^2 + 10x}{(x+3)(x+5)}
$$
- Combine the fractions:
$$
\frac{5x + 15}{(x+5)(x+3)} + \frac{2x^2 + 10x}{(x+3)(x+5)} = \frac{5x + 15 + 2x^2 + 10x}{(x+5)(x+3)}
$$
- Simplify the numerator:
$$
5x + 15 + 2x^2 + 10x = 2x^2 + 15x + 15
$$
- Final expression:
$$
\frac{2x^2 + 15x + 15}{(x+5)(x+3)}
$$
Solution:
$$
\boxed{\frac{2x^2 + 15x + 15}{(x+5)(x+3)}}
$$
---
$$
\frac{6}{x-5} + \frac{x+2}{x-5}
$$
- The denominators are the same: \( x-5 \).
- Combine the numerators:
$$
\frac{6 + (x+2)}{x-5} = \frac{6 + x + 2}{x-5} = \frac{x + 8}{x-5}
$$
Solution:
$$
\boxed{\frac{x+8}{x-5}}
$$
---
$$
\frac{4}{5b} - \frac{7a}{4b^2}
$$
- The denominators are \( 5b \) and \( 4b^2 \).
- Find the least common denominator (LCD): \( 20b^2 \).
- Rewrite each fraction with the LCD:
$$
\frac{4}{5b} = \frac{4 \cdot 4b}{5b \cdot 4b} = \frac{16b}{20b^2}
$$
$$
\frac{7a}{4b^2} = \frac{7a \cdot 5}{4b^2 \cdot 5} = \frac{35a}{20b^2}
$$
- Combine the fractions:
$$
\frac{16b}{20b^2} - \frac{35a}{20b^2} = \frac{16b - 35a}{20b^2}
$$
Solution:
$$
\boxed{\frac{16b - 35a}{20b^2}}
$$
---
$$
\frac{2x}{5x+4} + \frac{6x}{2x+3}
$$
- The denominators are \( 5x+4 \) and \( 2x+3 \).
- Find the least common denominator (LCD): \( (5x+4)(2x+3) \).
- Rewrite each fraction with the LCD:
$$
\frac{2x}{5x+4} = \frac{2x(2x+3)}{(5x+4)(2x+3)} = \frac{4x^2 + 6x}{(5x+4)(2x+3)}
$$
$$
\frac{6x}{2x+3} = \frac{6x(5x+4)}{(2x+3)(5x+4)} = \frac{30x^2 + 24x}{(2x+3)(5x+4)}
$$
- Combine the fractions:
$$
\frac{4x^2 + 6x}{(5x+4)(2x+3)} + \frac{30x^2 + 24x}{(5x+4)(2x+3)} = \frac{4x^2 + 6x + 30x^2 + 24x}{(5x+4)(2x+3)}
$$
- Simplify the numerator:
$$
4x^2 + 6x + 30x^2 + 24x = 34x^2 + 30x
$$
- Factor out the greatest common factor in the numerator:
$$
34x^2 + 30x = 2x(17x + 15)
$$
- Final expression:
$$
\frac{2x(17x + 15)}{(5x+4)(2x+3)}
$$
Solution:
$$
\boxed{\frac{2x(17x + 15)}{(5x+4)(2x+3)}}
$$
---
1. \(\boxed{\frac{3}{y-5}}\)
2. \(\boxed{\frac{6}{a+1}}\)
3. \(\boxed{\frac{4}{3x^2}}\)
4. \(\boxed{\frac{-5x^2 + 5x + 24}{4(x-1)}}\)
5. \(\boxed{\frac{2x^2 + 15x + 15}{(x+5)(x+3)}}\)
6. \(\boxed{\frac{x+8}{x-5}}\)
7. \(\boxed{\frac{16b - 35a}{20b^2}}\)
8. \(\boxed{\frac{2x(17x + 15)}{(5x+4)(2x+3)}}\)
1. Identify common denominators when adding or subtracting.
2. Combine numerators after ensuring the denominators are the same.
3. Simplify the resulting expression if possible.
Let's solve each problem step by step.
---
Problem 1:
$$
\frac{7}{y-5} - \frac{4}{y-5}
$$
- The denominators are the same: \( y-5 \).
- Combine the numerators:
$$
\frac{7 - 4}{y-5} = \frac{3}{y-5}
$$
Solution:
$$
\boxed{\frac{3}{y-5}}
$$
---
Problem 2:
$$
\frac{4}{a+1} + \frac{2}{a+1}
$$
- The denominators are the same: \( a+1 \).
- Combine the numerators:
$$
\frac{4 + 2}{a+1} = \frac{6}{a+1}
$$
Solution:
$$
\boxed{\frac{6}{a+1}}
$$
---
Problem 3:
$$
\frac{5}{3x^2} - \frac{1}{3x^2}
$$
- The denominators are the same: \( 3x^2 \).
- Combine the numerators:
$$
\frac{5 - 1}{3x^2} = \frac{4}{3x^2}
$$
Solution:
$$
\boxed{\frac{4}{3x^2}}
$$
---
Problem 4:
$$
\frac{6}{x-1} - \frac{5x}{4}
$$
- The denominators are different: \( x-1 \) and \( 4 \).
- Find the least common denominator (LCD): \( 4(x-1) \).
- Rewrite each fraction with the LCD:
$$
\frac{6}{x-1} = \frac{6 \cdot 4}{(x-1) \cdot 4} = \frac{24}{4(x-1)}
$$
$$
\frac{5x}{4} = \frac{5x \cdot (x-1)}{4 \cdot (x-1)} = \frac{5x(x-1)}{4(x-1)}
$$
- Combine the fractions:
$$
\frac{24}{4(x-1)} - \frac{5x(x-1)}{4(x-1)} = \frac{24 - 5x(x-1)}{4(x-1)}
$$
- Simplify the numerator:
$$
24 - 5x(x-1) = 24 - 5x^2 + 5x
$$
- Final expression:
$$
\frac{-5x^2 + 5x + 24}{4(x-1)}
$$
Solution:
$$
\boxed{\frac{-5x^2 + 5x + 24}{4(x-1)}}
$$
---
Problem 5:
$$
\frac{5}{x+5} + \frac{4x}{2x+6}
$$
- Simplify the second fraction:
$$
\frac{4x}{2x+6} = \frac{4x}{2(x+3)} = \frac{2x}{x+3}
$$
- The denominators are now \( x+5 \) and \( x+3 \).
- Find the least common denominator (LCD): \( (x+5)(x+3) \).
- Rewrite each fraction with the LCD:
$$
\frac{5}{x+5} = \frac{5(x+3)}{(x+5)(x+3)} = \frac{5x + 15}{(x+5)(x+3)}
$$
$$
\frac{2x}{x+3} = \frac{2x(x+5)}{(x+3)(x+5)} = \frac{2x^2 + 10x}{(x+3)(x+5)}
$$
- Combine the fractions:
$$
\frac{5x + 15}{(x+5)(x+3)} + \frac{2x^2 + 10x}{(x+3)(x+5)} = \frac{5x + 15 + 2x^2 + 10x}{(x+5)(x+3)}
$$
- Simplify the numerator:
$$
5x + 15 + 2x^2 + 10x = 2x^2 + 15x + 15
$$
- Final expression:
$$
\frac{2x^2 + 15x + 15}{(x+5)(x+3)}
$$
Solution:
$$
\boxed{\frac{2x^2 + 15x + 15}{(x+5)(x+3)}}
$$
---
Problem 6:
$$
\frac{6}{x-5} + \frac{x+2}{x-5}
$$
- The denominators are the same: \( x-5 \).
- Combine the numerators:
$$
\frac{6 + (x+2)}{x-5} = \frac{6 + x + 2}{x-5} = \frac{x + 8}{x-5}
$$
Solution:
$$
\boxed{\frac{x+8}{x-5}}
$$
---
Problem 7:
$$
\frac{4}{5b} - \frac{7a}{4b^2}
$$
- The denominators are \( 5b \) and \( 4b^2 \).
- Find the least common denominator (LCD): \( 20b^2 \).
- Rewrite each fraction with the LCD:
$$
\frac{4}{5b} = \frac{4 \cdot 4b}{5b \cdot 4b} = \frac{16b}{20b^2}
$$
$$
\frac{7a}{4b^2} = \frac{7a \cdot 5}{4b^2 \cdot 5} = \frac{35a}{20b^2}
$$
- Combine the fractions:
$$
\frac{16b}{20b^2} - \frac{35a}{20b^2} = \frac{16b - 35a}{20b^2}
$$
Solution:
$$
\boxed{\frac{16b - 35a}{20b^2}}
$$
---
Problem 8:
$$
\frac{2x}{5x+4} + \frac{6x}{2x+3}
$$
- The denominators are \( 5x+4 \) and \( 2x+3 \).
- Find the least common denominator (LCD): \( (5x+4)(2x+3) \).
- Rewrite each fraction with the LCD:
$$
\frac{2x}{5x+4} = \frac{2x(2x+3)}{(5x+4)(2x+3)} = \frac{4x^2 + 6x}{(5x+4)(2x+3)}
$$
$$
\frac{6x}{2x+3} = \frac{6x(5x+4)}{(2x+3)(5x+4)} = \frac{30x^2 + 24x}{(2x+3)(5x+4)}
$$
- Combine the fractions:
$$
\frac{4x^2 + 6x}{(5x+4)(2x+3)} + \frac{30x^2 + 24x}{(5x+4)(2x+3)} = \frac{4x^2 + 6x + 30x^2 + 24x}{(5x+4)(2x+3)}
$$
- Simplify the numerator:
$$
4x^2 + 6x + 30x^2 + 24x = 34x^2 + 30x
$$
- Factor out the greatest common factor in the numerator:
$$
34x^2 + 30x = 2x(17x + 15)
$$
- Final expression:
$$
\frac{2x(17x + 15)}{(5x+4)(2x+3)}
$$
Solution:
$$
\boxed{\frac{2x(17x + 15)}{(5x+4)(2x+3)}}
$$
---
Final Answers:
1. \(\boxed{\frac{3}{y-5}}\)
2. \(\boxed{\frac{6}{a+1}}\)
3. \(\boxed{\frac{4}{3x^2}}\)
4. \(\boxed{\frac{-5x^2 + 5x + 24}{4(x-1)}}\)
5. \(\boxed{\frac{2x^2 + 15x + 15}{(x+5)(x+3)}}\)
6. \(\boxed{\frac{x+8}{x-5}}\)
7. \(\boxed{\frac{16b - 35a}{20b^2}}\)
8. \(\boxed{\frac{2x(17x + 15)}{(5x+4)(2x+3)}}\)
Parent Tip: Review the logic above to help your child master the concept of adding and subtracting rational expressions worksheet pdf.