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Addition and Subtraction of Rational Expressions

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Adding and Subtracting Rational Expressions with Like Denominators
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In triangle ABC, side AB is \frac{x^2+5}{x^2 + 3x + 2} units long. Side AC is \frac{3x^2 - 3x}{x^2 + 3x + 2} units long. How much longer is side AC than side AB?

Guidance

Recall, that when you add or subtract fractions, the denominators must be the same. The same is true of adding and subtracting rational expressions. The denominators must be the same expression and then you can add or subtract the numerators.

Example A

Add \frac{x}{x-6} + \frac{7}{x-6} .

Solution: In this concept, the denominators will always be the same. Therefore, all you will need to do is add the numerators and simplify if needed.

\frac{x}{x-6} + \frac{7}{x-6} = \frac{x+7}{x-6}

Example B

Subtract \frac{x^2-4}{x-3} - \frac{2x-1}{x-3} .

Solution: You need to be a little more careful with subtraction. The entire expression in the second numerator is being subtracted. Think of the minus sign like distributing -1 to that numerator.

\frac{x^2-4}{x-3} - \frac{2x-1}{x-3} &= \frac{x^2-4-(2x+1)}{x-3} \\&= \frac{x^2-4-2x-1}{x-3} \\&= \frac{x^2-2x-3}{x-3}

At this point, factor the numerator if possible.

\frac{x^2-2x-3}{x-3} = \frac{\cancel{(x-3)}(x+1)}{\cancel{x-3}} = x+1

Example C

Add \frac{x+7}{2x^2+14x+20} + \frac{x+1}{2x^2+14x+20} .

Solution: Add the numerators and simplify the denominator.

\frac{x+7}{2x^2+14x+20} + \frac{x+1}{2x^2+14x+20} &= \frac{2x+8}{2x^2+14x+20} \\&= \frac{\cancel{2}(x+4)}{\cancel{2}(x+5)(x+2)} \\&= \frac{(x+4)}{(x+5)(x+2)}

Intro Problem Revisit We need to subtract the length of side AB from the length of side AC.

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

Now we need to factor the numerator and the denominator.

\frac{2x^2 - 3x - 5}{x^2 + 3x + 2} = \frac{(2x-5)(x+1)}{(x+2)(x+1)}

The (x + 1) in the numerator and the denominator cancel out and we are left with \frac{2x-5}{x+2} . Therefore, side AC is \frac{2x-5}{x+2} units longer than side AB.

Guided Practice

Add or subtract the following rational expressions.

1. \frac{3}{x^2-9} - \frac{x+7}{x^2-9}

2. \frac{5x-6}{2x+3} + \frac{x-12}{2x+3}

3. \frac{x^2+2}{4x^2-4x-3} - \frac{x^2-2x+1}{4x^2-4x-3}

Answers

1. \frac{3}{x^2-9} - \frac{x+7}{x^2-9} = \frac{3-(x+7)}{x^2-9} = \frac{3-x-7}{x^2-9} = \frac{-x-4}{x^2-9}

We did not bother to factor the denominator because we know that the factors of -9 are 3 and -3 and will not cancel with -x-4 .

2. \frac{5x-6}{2x+3} + \frac{x-12}{2x+3} = \frac{6x-18}{2x+3} = \frac{6(x-3)}{2x+3}

3. \frac{x^2+2}{4x^2-4x-3} - \frac{x^2-2x+1}{4x^2-4x-3} &= \frac{x^2+2-(x^2-2x+1)}{4x^2-4x-3} \\&= \frac{x^2+2-x^2+2x-1}{4x^2-4x-3} \\&= \frac{2x+1}{4x^2-4x-3}

At this point, we will factor the denominator to see if any factors cancel with the numerator.

\frac{2x+1}{4x^2-4x-3} = \frac{\cancel{2x+1}}{\cancel{(2x+1)}(2x-3)} = \frac{1}{2x-3}

Practice

  1. Explain how you add fractions. Assume your audience knows nothing about math.
  2. Explain why \frac{2}{3} + \frac{4}{5} \neq \frac{3}{4} .

Add or subtract the following rational expressions.

  1. \frac{2}{x} + \frac{5}{x}
  2. \frac{5}{2x} + \frac{7}{2x}
  3. \frac{6}{5x} + \frac{3-2x}{5x}
  4. \frac{3}{x} + \frac{x+1}{x}
  5. \frac{5}{x+1} + \frac{x-4}{x+1}
  6. \frac{x+15}{x-2} - \frac{10}{x-2}
  7. \frac{4x-3}{x+3} + \frac{15}{x+3}
  8. \frac{3x+8}{x^2-4x-5} + \frac{2x+3}{x^2-4x-5}
  9. \frac{5x+3}{x^2-4} - \frac{2x+9}{x^2-4}
  10. \frac{3x^2+x}{x^3-8} + \frac{4}{x^3-8} - \frac{2x^2-x}{x^3-8}
  11. \frac{4x+3}{x^2+1} - \frac{x+2}{x^2+1} + \frac{1-x}{x^2+1}
  12. \frac{18x^2-7x+2}{8x^3+4x^2-18x-9} - \frac{3x^2+13x-4}{8x^3+4x^2-18x-9} + \frac{5x^2-13}{8x^3+4x^2-18x-9}
  13. \frac{2x^2+3x}{x^3+2x^2-16x-32} + \frac{5x^2-13}{x^3+2x^2-16x-32} - \frac{4x^2+9x+11}{x^3+2x^2-16x-32}

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