9.3: Add and Subtract Rational Expressions
To add or subtract rational expressions , think of this as fractions with variables. A common denominator (called the LCD) is needed for addition and subtraction.
Find the LCD/LCM
To find the LCD , first fully factor all denominators. Build the LCD from the factors found in all denominators. Multiply each factor the greatest number of times it occurs in either expression. If the same factor occurs more than once in both expressions, multiply the factor the greatest number of times it occurs in either expression. It will be called the LCM in this section (Least Common Multiple), because there are no fractions in these problems.
- \((x^2 − 2x − 3)\) and \((x^2 + 2x − 15)\)
- \((x^2 − 9)\) and \((2x^2 − 5x − 3)\)
- \((x^2 + x − 2)\) and \((x^2 + 4x + 4)\)
Solution
- \(\begin{array} &&(x^2 − 2x − 3) \text{ and } (x^2 + 2x − 15) &\text{Example problem} \\ &(x − 3)(x + 1) \text{ and } (x − 3)(x + 5) &\text{Factor} \\ &\text{The LCM is } (x − 3)(x + 1)(x + 5) &\text{Final answer. Every factor must be represented in the answer.} \\& &\text{Only one copy of \((x − 3)\) is needed, because it represents the factor found in each expression.} \end{array}\)
- \(\begin{array} &&(x^2 − 9) \text{ and } (2x^2 − 5x − 3) &\text{Example problem} \\ &(x−3)(x+3) \text{ and } (2x^2−6x+1x−3) &\text{Factor; the first polynomial is a difference of squares, and use factor by grouping for the second polynomial.} \\ &(x−3)(x+3) \text{ and } (2x(x−3)+1(x− 3)) &\text{Factor by grouping.} \\ &(x − 3)(x + 3) \text{ and } (2x + 1)(x − 3) &\text{Completely factored.} \\ &\text{The LCM is } (x − 3)(x + 3)(2x + 1) &\text{Final answer. Every factor must be represented in the answer.} \\ & &\text{Only one copy of \((x − 3)\) is needed, because it represents the factor found in each expression.} \end{array}\)
- \(\begin{array} &&(x^2 + x − 2) \text{ and } (x^2 + 4x + 4) &\text{Example problem} \\ &(x − 1)(x + 2) \text{ and } (x + 2)(x + 2) &\text{Factor.} \\ &\text{The LCM is } (x − 1)(x + 2)(x + 2) &\text{Final answer. Every factor must be represented in the answer.} \\ & &\text{Two copies of \((x + 2)\) are needed, because it represents the greatest number of those factors found in either expression.} \\ &\text{The LCM is } (x − 1)(x + 2)^2 &\text{Alternate answer.} \end{array}\)
Find the LCM:
- \((3x^2 − 13x + 4)\) and \((x^2 − 16)\)
- \((2x^2 + x − 3)\) and \((x^2 − 2x + 1)\)
- \((x − 1)\) and \((x^2 − 4x − 5)\)
- \((6x^2 − 23x + 20)\) and \((4x^2 − 25)\)
Subtract Rational Expressions and Simplify to a Single Rational Expression
Rational expressions are fractions with variables (also known as algebraic fractions). To add or subtract rational expressions, first find the common denominator (the LCD), then add or subtract the numerators, keeping the same (common) denominator. Finally, factor and simplify by removing common factors from the numerator and denominator if possible.
Add or Subtract and Simplify:
- \(\dfrac{2x}{2x − 1} - \dfrac{2x}{2x + 5}\)
- \(\dfrac{4}{x^2 − 9} - \dfrac{5}{x^2 − 6x + 9}\)
- \(\dfrac{x}{1 + x} + \dfrac{2x + 3}{x^2 − 1}\)
Solution
- \(\begin{array} &&\dfrac{2x}{2x − 1} - \dfrac{2x}{2x + 5} &\text{Example problem} \\ &\dfrac{2x}{2x − 1} - \dfrac{2x}{2x + 5} &\text{Find the LCD, which is \((2x − 1)(2x + 5)\)} \\ &\dfrac{2x(2x + 5)}{(2x − 1)(2x + 5)} - \dfrac{2x(2x − 1)}{(2x − 1)(2x + 5)} &\text{Multiply the numerator and denominator of each rational expression by the missing terms in the LCD.} \\ &\dfrac{2x(2x + 5) − [2x(2x − 1)]}{(2x − 1)(2x + 5)} &\text{Place the subtraction in the numerator over a single common denominator.} \\ &\dfrac{4x^2 + 10x − [4x^2 − 2x]}{(2x − 1)(2x + 5)} &\text{Distribute, combine like terms and simplify the numerator.} \\ &\dfrac{4x^2 + 10x − 4x^2 + 2x}{(2x − 1)(2x + 5)} &\text{Take care to distribute the subtraction to both terms.} \\ &\dfrac{12x}{(2x − 1)(2x + 5)} &\text{Final answer.} \end{array}\)
- \(\begin{array} &&\dfrac{4}{x^2 − 9} - \dfrac{5}{x^2 − 6x + 9} &\text{Example problem} \\ &\dfrac{4}{(x + 3)(x − 3)} - \dfrac{5}{(x − 3)(x − 3)} &\text{Factor the denominators.} \\ &\dfrac{4}{(x + 3)(x − 3)} - \dfrac{5}{(x − 3)(x − 3)} &\text{Find the LCD, which is \((x − 3)(x − 3)(x + 3)\)} \\ &\dfrac{4(x − 3)}{(x + 3)(x − 3)(x − 3)} - \dfrac{5(x + 3)}{(x − 3)(x − 3)(x + 3)} &\text{Multiply the numerator and denominator of each rational expression by the missing terms in the LCD.} \\ &\dfrac{4}{(x − 3) − 5(x + 3)}{(x + 3)(x − 3)(x − 3)} &\text{Place the subtraction in the numerator over a single common denominator.} \\ &\dfrac{4x − 12 − [5x + 15]}{(x + 3)(x − 3)(x − 3)} &\text{Distribute, combine like terms and simplify the numerator.} \\ &\dfrac{4x − 12 − 5x − 15}{(x + 3)(x − 3)(x − 3)} &\text{Take care to distribute the subtraction to both terms.} \\ &\dfrac{−x − 27}{(x + 3)(x − 3)(x − 3)} &\text{Final answer.} \\ &\dfrac{−(x + 27)}{(x + 3)(x − 3)(x − 3)} &\text{Alternate answer} \end{array}\)
- \(\begin{array} &&\dfrac{x}{1 + x} + \dfrac{2x + 3}{x^2 − 1} &\text{Example problem} \\ &\dfrac{x}{x + 1} + \dfrac{2x + 3}{(x − 1)(x + 1)} &\text{Factor the denominators.} \\ &\dfrac{x}{x + 1} + \dfrac{2x + 3}{(x − 1)(x + 1)} &\text{Find the LCD, which is \((x − 1)(x + 1)\)} \\ &\dfrac{x(x − 1)}{(x + 1)(x − 1)} + \dfrac{(2x + 3)}{(x − 1)(x + 1)} &\text{Multiply the numerator and denominator of each rational expression by the missing terms in the LCD.} \\ & &\text{Notice that the second rational expression already has the LCD as its denominator.} \\[0.125in] &\dfrac{x(x − 1) + (2x + 3)}{(x + 1)(x − 1)} &\text{Place the subtraction in the numerator over a single common denominator.} \\ &\dfrac{x^2 − x + 2x + 3}{(x + 1)(x − 1)} &\text{Distribute, combine like terms and simplify the numerator.} \\ &\dfrac{x^2 + x + 3}{(x + 1)(x − 1)} &\text{Take care to distribute the subtraction to both terms.} \\ &\dfrac{x^2 + x + 3}{(x + 1)(x − 1)} &\text{Final answer.} \end{array}\)
Add or Subtract and Simplify:
- \(\dfrac{x}{x^2 + 1} + \dfrac{24x^3}{x3 + 2}\)
- \(\dfrac{x}{1 − x} + \dfrac{2x + 3}{x^2 − 1}\)
- \(\dfrac{5}{x + 3} + \dfrac{x^2 − 4x − 21}{x^2 − 9}\)
- \(\dfrac{39x + 36}{x^2 − 3x − 10} - \dfrac{23x − 16}{x^2 − 7x + 10}\)