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Adding and Subtracting Rational Expressions where One Denominator is the LCD

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Adding and Subtracting Rational Expressions where One Denominator is the LCD

The length of a garden plot is \frac{6x^2-5}{2x^2 + 4x - 6} . The width of the plot is \frac{2x-7}{x+3} . How much longer is the garden plot than it is wide?

Guidance

Recall when two fractions do not have the same denominator. You have to multiply one or both fractions by a number to create equivalent fractions in order to combine them.

\frac{1}{2} + \frac{3}{4}

Here, 2 goes into 4 twice. So, we will multiply the first fraction by {\color{red}\frac{2}{2}} to get a denominator of 4. Then, the two fractions can be added.

{\color{red}\frac{2}{2} \cdot} \frac{1}{2} + \frac{3}{4} = \frac{2}{4} + \frac{3}{4} = \frac{5}{4}

Once the denominators are the same, the fractions can be combined. We will apply this idea to rational expressions in order to add or subtract ones without like denominators.

Example A

Subtract \frac{3x-5}{2x+8} - \frac{x^2-6}{x+4} .

Solution: Factoring the denominator of the first fraction, we have 2(x+4) . The second fraction needs to be multiplied by {\color{red}\frac{2}{2}} in order to make the denominators the same.

\frac{3x-5}{2x+8} - \frac{x^2-6}{x+4} &= \frac{3x-5}{2(x+4)} - \frac{x^2-6}{x+4} \cdot {\color{red}\frac{2}{2}} \\&= \frac{3x-5}{2(x+4)} - \frac{2x^2-12}{2(x+4)}

Now that the denominators are the same, subtract the second rational expression just like in the previous concept.

&= \frac{3x-5-(2x^2-12)}{2(x+4)} \\&= \frac{3x-5-2x^2+12}{2(x+4)} \\&= \frac{-2x^2+3x+7}{2(x+4)}

The numerator is not factorable, so we are done.

Example B

Add \frac{2x-3}{x+5} + \frac{x^2+1}{x^2-2x-35}

Solution: Factoring the second denominator, we have x^2-2x-35=(x+5)(x-7) . So, we need to multiply the first fraction by {\color{red}\frac{x-7}{x-7}} .

\overbrace{{\color{red}\frac{(x-7)}{(x-7)}} \cdot \frac{(2x-3)}{(x+5)}}^{FOIL} + \frac{x^2+1}{(x-7)(x+5)} &= \frac{2x^2-17x+21}{(x-7)(x+5)} + \frac{x^2+1}{(x-7)(x+5)} \\&= \frac{3x^2-17x+22}{(x-7)(x+5)}

Example C

Subtract \frac{7x+2}{2x^2+18x+40} - \frac{6}{x+5} .

Solution: Factoring the first denominator, we have 2x^2+18x+40=2(x^2+9x+20)=2(x+4)(x+5) . This is the Lowest Common Denominator, or LCD. The second fraction needs the 2 and the (x+4) .

\frac{7x+2}{2x^2+18x+40} - \frac{6-x}{x+5} &= \frac{7x+2}{2(x+5)(x+4)} - \frac{6-x}{x+5}{\color{red}\cdot \frac{2(x+4)}{2(x+4)}} \\&= \frac{7x+2}{2(x+5)(x+4)} - \frac{2(6-x)(x+4)}{2(x+5)(x+4)} \\&= \frac{7x+2}{2(x+5)(x+4)} - \frac{48+4x-2x^2}{2(x+5)(x+4)} \\&= \frac{7x+2-(48+4x-2x^2)}{2(x+5)(x+4)} \\&= \frac{7x+2-48-4x+2x^2}{2(x+5)(x+4)} \\&= \frac{2x^2+3x-46}{2(x+5)(x+4)}

Intro Problem Revisit We need to subtract the width from the length.

\frac{6x^2-5}{2x^2 + 4x - 6} - \frac{2x-7}{x+3}

Factoring the first denominator, we have 2x^2+4x-6=(2x-2)(x+3) . So, we need to multiply the second fraction by {\color{red}\frac{2x-2}{2x-2}} .

\frac{6x^2-5}{2x^2 + 4x - 6} -\overbrace{{\color{red}\frac{(2x-2)}{(2x-2)}} \cdot \frac{(2x-7)}{(x+3)}}^{FOIL} &= \frac{4x^2-18x+14}{2x^2+4x-6} \\\frac{6x^2-5}{2x^2 + 4x - 6} - \frac{4x^2-18x+14}{2x^2+4x-6}\\\frac{6x^2-5-(4x^2-18x+14)}{2x^2+4x-6}\\\frac{6x^2-5-4x^2+18x-14}{2x^2+4x-6}\\\frac{2x^2+18x-19}{2x^2+4x-6}

Therefore, the garden plot is \frac{2x^2+18x-19}{2x^2+4x-6} longer than it is wide.

Guided Practice

Perform the indicated operation.

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

2. \frac{x-10}{x^2+4x-24} + \frac{x+3}{x+6}

3. \frac{3x^2-5}{3x^2-12} + \frac{x+8}{3x+6}

Answers

1. The LCD is 3x+3 or 3(x+1) . Multiply the first fraction by \frac{3}{3} .

\frac{2}{x+1} - \frac{x}{3x+3} &= \frac{3}{3} \cdot \frac{2}{x+1} - \frac{x}{3(x+1)} \\&= \frac{6}{3(x+1)} - \frac{x}{3(x+1)} \\&= \frac{6-x}{3(x+1)}

2. Here, the LCD x^2+4x-24 or (x+6)(x-4) . Multiply the second fraction by \frac{x-4}{x-4} .

\frac{x-10}{x^2+4x-24} + \frac{x+3}{x+6} &= \frac{x-10}{(x+6)(x-4)} + \frac{x+3}{x+6} \cdot \frac{x-4}{x-4} \\&= \frac{x-10}{(x+6)(x-4)} + \frac{x^2-x-12}{(x+6)(x-4)} \\&= \frac{x-10+x^2-x-12}{(x+6)(x-4)} \\&= \frac{x^2-22}{(x+6)(x-4)}

3. The LCD is 3x^2-12=3(x-2)(x+2) . The second fraction’s denominator factors to be 3x+6=3(x+2) , so it needs to be multiplied by \frac{x-2}{x-2} .

\frac{3x^2-5}{3x^2-12} + \frac{x+8}{3x+6} &= \frac{3x^2-5}{3(x-2)(x+2)} + \frac{x+8}{3(x+2)} \cdot \frac{x-2}{x-2} \\&= \frac{3x^2-5}{3(x-2)(x+2)} + \frac{x^2+6x-16}{3(x-2)(x+2)} \\&= \frac{3x^2-5+x^2+6x-16}{3(x-2)(x+2)} \\&= \frac{4x^2+6x-21}{3(x-2)(x+2)}

Practice

Find the LCD.

  1. x, \ 6x
  2. x, \ x+1
  3. x+2, \ x-4
  4. x, \ x-1, \ x^2 - 1

Perform the indicated operations.

  1. \frac{3}{x} - \frac{5}{4x}
  2. \frac{x+2}{x+3} + \frac{x-1}{x^2+3x}
  3. \frac{x}{x-7} - \frac{2x+7}{3x-21}
  4. \frac{x^2+3x-10}{x^2-4} - \frac{x}{x+2}
  5. \frac{5x+14}{2x^2-7x-15} - \frac{3}{x-5}
  6. \frac{x-3}{3x^2+x-10} + \frac{3}{x+2}
  7. \frac{x+1}{6x+2} + \frac{x^2-7x}{12x^2-14x-6}
  8. \frac{-3x^2-10x+15}{10x^2-x-3} + \frac{x+4}{2x+1}
  9. \frac{8}{2x-5} - \frac{x+5}{2x^2+x-15}
  10. \frac{2}{x+2} + \frac{3x+16}{x^2-x-6} - \frac{2}{x-3}
  11. \frac{6x^2+4x+8}{x^3+3x^2-x-3} + \frac{x-4}{x^2-1} - \frac{3x}{x^2+2x-3}

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