How do you simplify rational expressions? The rules of addition, subtraction, multiplication, and division of rational expressions are used to simplify complex algebraic fractions. Grade 11 examples are presented below along with detailed, step-by-step solutions.
Tip: You can use our online calculator to simplify rational expressions to check your final results.
If you have difficulties, review the tutorials on Addition & Subtraction and Multiplication & Division before starting.
| Addition & Subtraction (Requires a common denominator) |
\[ \dfrac{A}{B} \pm \dfrac{C}{B} = \dfrac{A \pm C}{B} \] |
| Multiplication | \[ \dfrac{A}{B} \cdot \dfrac{C}{D} = \dfrac{A \cdot C}{B \cdot D} \] |
| Division (Multiply by the reciprocal) |
\[ \dfrac{A}{B} \div \dfrac{C}{D} = \dfrac{\dfrac{A}{B}}{\dfrac{C}{D}} = \dfrac{A}{B} \cdot \dfrac{D}{C} = \dfrac{A \cdot D}{B \cdot C} \] |
Simplify: \[ \dfrac{\dfrac{2}{3} + 5}{4} +\dfrac{1}{2} \]
We first simplify the numerator \( \dfrac{2}{3} + 5 \). Convert to a common denominator.
\[ \dfrac{\dfrac{2}{3} + 5}{4} +\dfrac{1}{2} = \dfrac{\dfrac{2}{3} + 5 \cdot \dfrac{3}{3}}{4} +\dfrac{1}{2} = \dfrac{\dfrac{17}{3}}{4} +\dfrac{1}{2} \]
Change 4 to a fraction \( \dfrac{4}{1} \) and simplify by dividing the fractions (multiplying by the reciprocal).
\[ = \dfrac{\dfrac{17}{3}}{\dfrac{4}{1}} +\dfrac{1}{2} = \dfrac{17}{3} \cdot \dfrac{1}{4} +\dfrac{1}{2} \]
Multiply the fractions.
\[ = \dfrac{17 \cdot 1}{3 \cdot 4} + \dfrac{1}{2} = \dfrac{17}{12} +\dfrac{1}{2} \]
Convert to a common denominator and add.
\[ = \dfrac{17}{12} +\dfrac{1}{2} \cdot \dfrac{6}{6} = \dfrac{17+6}{12} = \mathbf{\dfrac{23}{12}} \]
Simplify: \[ \dfrac{\dfrac{x+1}{x-2}+\dfrac{x}{x+1}}{\dfrac{1}{x+1}} \]
We first simplify the numerator by converting to the lowest common denominator \( (x - 2)(x + 1) \).
\[ = \dfrac{\dfrac{x+1}{x-2} \cdot \dfrac{x+1}{x+1} +\dfrac{x}{x+1} \cdot \dfrac{x-2}{x-2} }{\dfrac{1}{x+1}} \]
\[ = \dfrac{ \dfrac{(x+1)^2+x(x-2)}{(x+1)(x-2)} }{\dfrac{1}{x+1}}\]
Apply the rule of division by multiplying by the reciprocal of the denominator.
\[ = \dfrac{((x+1)^2+x(x-2))}{(x+1)(x-2)} \cdot \dfrac{x+1}{1} \]
Simplify by canceling common factors.
\[ = \dfrac{((x+1)^2+x(x-2))}{{\cancel{(x+1)}}(x-2)} \cdot \dfrac{\cancel{(x+1)}}{1} = \dfrac{(x+1)^2+x(x-2)}{x-2}\]
Expand and simplify the numerator.
\[ = \dfrac{(x^2+2x+1) + (x^2-2x)}{x-2} = \mathbf{\dfrac{2x^2+1}{x-2}} \;\; \text{for} \;\; x \ne -1\]
Simplify: \[ \dfrac{\dfrac{x-1}{3x+2}+3}{\dfrac{x+1}{6x+4}} - 2 \]
Convert the numerator's terms to the same denominator.
\[ \dfrac{\dfrac{x-1}{3x+2}+3 \cdot \dfrac{3x+2}{3x+2}}{\dfrac{x+1}{6x+4}} - 2 = \dfrac{\dfrac{x - 1 +3(3x+2)}{3x+2}}{\dfrac{x+1}{6x+4}} - 2 \]
Expand and simplify the top.
\[ = \dfrac{\dfrac{10x + 5}{3x+2}}{\dfrac{x+1}{6x+4}} - 2 \]
Multiply by the reciprocal.
\[ = \dfrac{10x + 5}{3x+2} \cdot \dfrac{6x+4}{x+1} - 2 \]
Factor out constants to find common cancellations.
\[ = \dfrac{5(2x + 1)}{3x+2} \cdot \dfrac {2(3x+2)}{x+1} - 2 \]
Cancel \( (3x+2) \) and multiply.
\[ = \dfrac{5(2x + 1)}{{\cancel{3x+2}}} \cdot \dfrac {2{\cancel{(3x+2)}}}{x+1} - 2 = \dfrac{10 (2x+1)}{x+1}-2 \]
Convert to the same denominator and combine.
\[ = \dfrac{10(2x+1)}{x+1} - 2 \cdot \dfrac{x+1}{x+1} = \dfrac{20x + 10 - 2x - 2}{x+1} = \dfrac{18x+8}{x+1} = \mathbf{\dfrac{2(9x+4)}{x+1}} \]
Simplify: \[ \dfrac{1}{1+\dfrac{1}{x+\dfrac{1}{x+1}}} \]
Start from the deepest fraction and work upwards. Convert the terms in \( x + \dfrac{1}{x+1} \) to the same denominator.
\[ \dfrac{1}{1+\dfrac{1}{x \cdot \dfrac{x+1}{x+1}+\dfrac{1}{x+1}}} = \dfrac{1}{1+\dfrac{1}{\dfrac{x^2+x+1}{x+1}}} \]
Note that \( \dfrac{1}{\dfrac{A}{B}} \) is the reciprocal \( \dfrac{B}{A} \).
\[ = \dfrac{1}{1+ \dfrac{x+1}{x^2+x+1}} \]
Convert the remaining denominator terms to a common denominator.
\[ = \dfrac{1}{1 \cdot \dfrac{x^2+x+1}{x^2+x+1}+ \dfrac{x+1}{x^2+x+1}} \]
Simplify and add.
\[ = \dfrac{1}{ \dfrac{x^2+x+1+x+1}{x^2+x+1}} = \dfrac{1}{ \dfrac{x^2+2x+2}{x^2+x+1}}\]
Take the final reciprocal.
\[ = \mathbf{\dfrac{x^2+x+1}{x^2+2x+2}} \]
Simplify the following expressions. Click on "Show Solution" to verify your answers.
Problem: \( \dfrac{\dfrac{2}{5} + 7}{\dfrac{4}{3}} +\dfrac{1}{3} \)
Answer: \( \dfrac{353}{60} \)
Problem: \( \dfrac{\dfrac{x-2}{x+3}+\dfrac{x}{x+3}}{\dfrac{1}{2x+6}} \)
Answer: \( 4(x-1) \)
Problem: \( \dfrac{\dfrac{x-1}{3x-12}+3}{\dfrac{x+1}{x-4}} - \dfrac{2}{3} \)
Answer: \( \dfrac{8x-39}{3(x+1)} \)
Problem: \( \dfrac{ \dfrac{2-x}{x+1}-\dfrac{4}{x+1}}{\dfrac{1}{x-5}-\dfrac{4}{x-5}} \)
Answer: \( \dfrac{ (x+2)(x-5)}{3(x+1)} \)
Problem: \( \dfrac{ \dfrac{2-x}{x+2} - \dfrac{4}{x+3}}{\dfrac{1}{x+3} \cdot \dfrac{4}{x+3}} \)
Answer: \( \dfrac{(-x^2-5x-2)(x+3)}{4(x+2)} \)
Problem: \( \dfrac{\dfrac{1}{2}}{1+\dfrac{1}{\dfrac{1}{x-1}-2}} \)
Answer: \( \dfrac{-2x+3}{2(-x+2)} \)
Challenge 1: Simplify a complex fraction requiring polynomial factoring.
\[ \dfrac{ \dfrac{x^2-9}{x^2-4x+4} }{ \dfrac{x+3}{x-2} } \]
Step 1: Apply the division rule (multiply by the reciprocal).
\[ = \dfrac{x^2-9}{x^2-4x+4} \cdot \dfrac{x-2}{x+3} \]
Step 2: Factor the numerators and denominators. Notice the difference of squares and the perfect square trinomial.
\[ = \dfrac{(x-3)(x+3)}{(x-2)(x-2)} \cdot \dfrac{x-2}{x+3} \]
Step 3: Cancel the common factors \( (x+3) \) and \( (x-2) \).
\[ = \dfrac{(x-3){\cancel{(x+3)}}}{(x-2){\cancel{(x-2)}}} \cdot \dfrac{\cancel{x-2}}{\cancel{x+3}} \]
Answer: \( \mathbf{\dfrac{x-3}{x-2}} \)
Challenge 2: Add and subtract rational expressions with quadratic denominators.
\[ \dfrac{2x}{x^2-1} - \dfrac{1}{x-1} + \dfrac{3}{x+1} \]
Step 1: Factor all denominators to find the Least Common Denominator (LCD). Here, \( x^2-1 = (x-1)(x+1) \).
The LCD is \( (x-1)(x+1) \).
Step 2: Rewrite each fraction with the LCD.
\[ \dfrac{2x}{(x-1)(x+1)} - \dfrac{1 \cdot (x+1)}{(x-1)(x+1)} + \dfrac{3 \cdot (x-1)}{(x-1)(x+1)} \]
Step 3: Combine over the single denominator. Be careful to distribute the negative sign!
\[ = \dfrac{2x - (x+1) + 3(x-1)}{(x-1)(x+1)} \]
Step 4: Expand and combine like terms in the numerator.
\[ = \dfrac{2x - x - 1 + 3x - 3}{(x-1)(x+1)} = \dfrac{4x - 4}{(x-1)(x+1)} \]
Step 5: Factor the numerator to see if anything else cancels.
\[ = \dfrac{4(x-1)}{(x-1)(x+1)} \]
Answer: \( \mathbf{\dfrac{4}{x+1}} \)
Challenge 3: Simplify a multi-level continued fraction.
\[ 2 - \dfrac{1}{2 - \dfrac{1}{2 - x}} \]
Step 1: Simplify the innermost denominator \( 2 - \dfrac{1}{2 - x} \).
\[ 2 \cdot \dfrac{2-x}{2-x} - \dfrac{1}{2-x} = \dfrac{4 - 2x - 1}{2-x} = \dfrac{3 - 2x}{2-x} \]
Step 2: Substitute this back into the main equation. Remember that \( \dfrac{1}{\frac{A}{B}} = \dfrac{B}{A} \).
\[ 2 - \dfrac{1}{\dfrac{3 - 2x}{2-x}} = 2 - \dfrac{2-x}{3-2x} \]
Step 3: Find a common denominator to combine the final terms.
\[ = \dfrac{2(3-2x)}{3-2x} - \dfrac{2-x}{3-2x} = \dfrac{6 - 4x - (2 - x)}{3-2x} \]
Step 4: Combine like terms.
\[ = \dfrac{6 - 4x - 2 + x}{3-2x} \]
Answer: \( \mathbf{\dfrac{4 - 3x}{3 - 2x}} \)