Algebra I

Solving Rational Expressions

Big Picture

Rational expressions are fractions that contain variables instead of just integers in the numerator and denominator. Just like fractions, rational expressions are a result of dividing the numerator by the denominator. Operations such as addition, subtraction, and multiplication can be performed on rational expressions as well. You can also use tools that apply to other equations, such as commutative, distributive, and associative properties. Equations involving rational expressions can be simplified by finding the least common denominators, making them easier to solve. It is important to follow the order of operations, even though it is a fraction.

Key Terms

Rational Expression: An expression that is the ratio of two polynomials and the denominator is not \(0\).

Excluded Value: A value that makes a rational expression undefined (\(denominator = 0\)).

Rational Equation: An equation with one or more rational expressions.

Simplifying Rational Expressions

Rational expressions are basically the quotient (result of division) of two polynomials.

  • To divide a polynomial by a monomial, we can divide each term in the numerator by the monomial.
  • The quotient rule for exponents is. \(\frac{x^n}{x^m} = x^{n-m}\).
  • To divide a polynomial by a binomial, use long division.
  • Remember that:
    Dividend \(\div\) Divisor \(=\) Quotient \(+ \frac{Remainder}{Divisor}\)
  • The dividend is the numerator, and the divisor is the denominator.

When we simplify rational expressions, we want the numerator and denominator to not have any factors in common.

  • Factor first, then cancel common factors.
  • Make sure to cancel common factors from thedenominator, not common terms!
  • Keep an eye out for two terms like \(x-1\) and \(1-x\).These terms are actually opposites!
  • \(1-x = -(x-1)\).
  • If we had the expression \(\frac{x - 1}{1 - x'}\) we can simplify it so that \(\frac{x - 1}{-(x - 1)} = -1\)

Excluded Values

Excluded values are the values that make the denominator zero. We cannot divide by zero, so we exclude any valuesthat cause the denominator to become zero.

  • Find excluded values by setting the denominator equal to \(0\).
  • Excluded values are related to vertical asymptotes.

Removable Zeros

When we cancel common factors, we may remove a division by zero.

Example: \(\frac{4x - 2}{2x^2 + x -1}\)

The denominator equals \(0\) when \(x = \frac{1}{2}\) or \(x = -1\)

\(\frac{4x - 2}{2x^2 + x - 1} = \frac{2(2x - 1)}{(2x - 1)(x + 1)}\)

If we cancel common factors first, we get \(\frac{2}{x + 1}\)

Now the denominator equals \(0\) when \(x = 1\). We have removed \(x = \frac{1}{2}\) as an excluded value.

Technically the original expression and the simplified one are not the same. We should write:

\(\frac{4x - 2}{2x^2 + x - 1} = \frac{2}{x + 1}, x \neq \frac{1}{2}\)

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Algebra I

Solving Rational Expressions cont.

Operations

Multiplication and Division

  • Factor and cancel out commonfactors before solving.
  • When dividing, convert the problem into a multiplication problem by flipping the second fraction, then simplify

Let \(a, b, c,\) and \(d\) be polynomials.

  • \(\frac{a}{b} . \frac{c}{d} = \frac{ac}{bd}, b \neq 0\) and \(d \neq 0\)
  • \(\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} . \frac{d}{c} = \frac{ad}{bc}, b \neq 0, c \neq 0\) and \(d \neq 0\)

Addition and Subtraction

Sums and differences with the same denominator:

  • \(\frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}\) and \(\frac{a}{c} - \frac{b}{c} = \frac{a - b}{c}\)

Sums and differences with different denominators:

  • Find the least common denominator (LCD) of the fractions.
  • Rewrite each fraction with the LCD as the denominator.
  • Add or subtract and simplify the result.

To find the LCD, we find the least common multiple of the denominators of the different fractions. The least common multiple of two or more integers is the least positive integer that has all of those integers as factors.

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

The denominators cannot be factored, so the LCD is the product of the denominators: \((x+2)(2x-5)\).

\( \frac{2}{x + 2} . \frac{(2x - 5)}{(2x - 5)} - \frac{3}{2x - 5} . \frac{(x + 2)}{(x + 2)}\)

\( \frac{2(2x - 5) - 3(x + 2)}{(x + 2)(2x - 5)} = \frac{4x - 10 - 3x - 6}{(x + 2)(2x - 5)} = \frac{x - 16}{(x + 2)(2x - 5)}\)

Solving Rational Equations

Rational equations with one term on each side:

  • Solve by cross-multiplication. Multiply each numerator by the denominator from the opposite side of the equation

    Example: \(\frac{x}{5}\)

    \(=\)

    \(\frac{x + 1}{2}\)

    \(\frac{x}{5} . 5\)

    \(=\)

    \(\frac{x + 1}{2} . 5\)

    \(x . 5\)

    \(=\)

    \(\frac{5(x + 1)}{2} . 2\)

    \(2x\)

    \(=\)

    \(5(x + 1)\)

    \(2x\)

    \(=\)

    \(5x + 5\)

    \(-5\)

    \(=\)

    \(3x\)

    \(x\)

    \(=\)

    \(- \frac{3}{5}\)

    We can use lowest common denominators to solve rational equations.

    • First find the LCD of the rational expressions.
    • Multiply by the LCD on both sides to simplify the equation.

    Example: \(\frac{x}{x - 2} + \frac{1}{5} = \frac{2}{x - 2}\)

    The LCD is 5(x-2), so multiply the original equation by the LCD.

    \(\frac{x}{x - 2} . 5(x - 2) + \frac{1}{5} . 5(x - 2)\)

    \(=\)

    \(\frac{2}{x - 2} . 5(x - 2)\)

    \(5x + (x - 2)\)

    \(=\)

    \(2 . 5\)

    \(6x - 2\)

    \(=\)

    \(10\)

    \(x\)

    \(=\)

    \(\frac{12}{6} = 2\)

    Be careful of extraneous solutions! \(x = 2\) looks like a solution, but this would make \(\frac{x}{x - 2}\) and \(\frac{2}{x - 2}\) undefined.

    Since \(x = 2\) is actually an extraneous solution, this rational equation has no solution!

    Notes

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