Holt McDougal Algebra 2 8-3 Adding and Subtracting Rational Expressions Add and subtract rational expressions. Simplify complex fractions. Objectives

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Holt McDougal Algebra 2

8-3 Adding and Subtracting Rational Expressions

Add and subtract rational expressions.

Simplify complex fractions.

Objectives



Adding and subtracting rational expressions is similar to adding and subtracting fractions. To add or subtract rational expressions with like denominators, add or subtract the numerators and use the same denominator.



Add or subtract. Identify any x-values for which the expression is undefined.

Example 1A: Adding and Subtracting Rational Expressions with Like Denominators

Add the numerators.

The expression is undefined at x = –4 because this value makes x + 4 equal 0.

x – 3 x + 4

+ x – 2x + 4

x – 3 +x + 4

x – 2

2x – 5x + 4

Combine like terms.




Example 1B: Adding and Subtracting Rational Expressions with Like Denominators

Subtract the numerators.

There is no real value of x for which x2 + 1 = 0; the expression is always defined.

3x – 4 x2 + 1

– 6x + 1x2 + 1

3x – 4 –x2 + 1 (6x + 1)

–3x – 5x2 + 1 Combine like terms.

Distribute the negative sign. 3x – 4 –

x2 + 1 6x – 1



Check It Out! Example 1a


Add the numerators.

6x + 5 x2 – 3

+ 3x – 1x2 – 3

6x + 5 +x2 – 3

3x – 1

9x + 4x2 – 3

Combine like terms.

The expression is undefined at x = ± because this value makes x2 – 3 equal 0.



To add or subtract rational expressions with unlike denominators, first find the least common denominator (LCD). The LCD is the least common multiple of the polynomials in the denominators.



Add. Identify any x-values for which the expression is undefined.

Example 3A: Adding Rational Expressions

x – 3 x2 + 3x – 4

+ 2x x + 4

x – 3 (x + 4)(x – 1)

+ 2x x + 4 Factor the denominators.

The LCD is (x + 4)(x – 1), so multiply by . 2x

x + 4

x – 1 x – 1

x – 3 (x + 4)(x – 1)

+ 2x x + 4

x – 1 x – 1



x – 3 + 2x(x – 1) (x + 4)(x – 1)

Add the numerators.

2x2 – x – 3 (x + 4)(x – 1)

Simplify the numerator.

Example 3A Continued


Write the sum in factored or expanded form.

2x2 – x – 3 (x + 4)(x – 1)

2x2 – x – 3 x2 + 3x – 4

or

The expression is undefined at x = –4 and x = 1 because these values of x make the factors (x + 4) and (x – 1) equal 0.




x x + 2

+ –8 x2 – 4

x x + 2

+ –8(x + 2)(x – 2) Factor the denominator.

The LCD is (x + 2)(x – 2), so multiply by . x

x + 2

x – 2 x – 2

Example 3B: Adding Rational Expressions

x – 2 x – 2

x x + 2

+ –8(x + 2)(x – 2)

x(x – 2) + (–8) (x + 2)(x – 2)

Add the numerators.



x2 – 2x – 8 (x + 2)(x – 2)

Write the numerator in standard form.


Example 3B Continued

(x + 2)(x – 4)(x + 2)(x – 2)

Factor the numerator.

x – 4x – 2

Divide out common factors.

The expression is undefined at x = –2 and x = 2 because these values of x make the factors (x + 2) and (x – 2) equal 0.





3x 2x – 2

+ 3x – 2 3x – 3

Factor the denominators.

The LCD is 6(x – 1), so

multiply by 3 and

by 2.

3x 2(x – 1)

3x – 2 3(x – 1)

3x 2(x – 1)

+ 3x – 2 3(x – 1)

3 3

3x 2(x – 1)

+ 3x – 2 3(x – 1)

2 2



Check It Out! Example 3a Continued


15x – 4 6(x – 1)

Simplify the numerator.

The expression is undefined at x = 1 because this value of x make the factor (x – 1) equal 0.

Add the numerators.9x + 6x – 4 6(x – 1)




2x + 6 x2 + 6x + 9

+x x + 3

Factor the denominators.

Check It Out! Example 3b

2x + 6 (x + 3)(x + 3)

+x x + 3

The LCD is (x + 3)(x + 3), so

multiply by . x

(x + 3) (x + 3) (x + 3)

2x + 6 (x + 3)(x + 3)

+x x + 3

x + 3 x + 3

Add the numerators.x2 + 3x + 2x + 6 (x + 3)(x + 3)



Check It Out! Example 3b Continued


Write the numerator in standard form.

(x + 3)(x + 2)(x + 3)(x + 3)


x + 2x + 3


The expression is undefined at x = –3 because this value of x make the factors (x + 3) and (x + 3) equal 0.

x2 + 5x + 6 (x + 3)(x + 3)



Subtract . Identify any x-

values for which the expression is undefined.

3x – 22x + 5

–5x – 2 2

(3x – 2)(5x – 2) – 2(2x + 5)(2x + 5)(5x – 2)

Subtract the numerators.

15x2 – 16x + 4 – (4x + 10)(2x + 5)(5x – 2)

Multiply the binomials in the numerator.


3x – 22x + 5

–5x – 2 2 2x + 5

2x + 5 5x – 2 5x – 2

The LCD is (2x + 5)(5x – 2),

so multiply by

and by .

3x – 2 2x + 5

(5x – 2) (5x – 2)

2 5x – 2

(2x + 5) (2x + 5)




Distribute the negative sign.

The expression is undefined at x = – and x = because these values of x make the factors (2x + 5) and (5x – 2) equal 0.

15x2 – 16x + 4 – 4x – 10(2x + 5)(5x – 2)

5 2

2 5

Subtract . Identify any x-

values for which the expression is undefined.

3x – 22x + 5

–5x – 2 2



Some rational expressions are complex fractions. A complex fraction contains one or more fractions in its numerator, its denominator, or both. Examples of complex fractions are shown below.

Recall that the bar in a fraction represents division. Therefore, you can rewrite a complex fraction as a division problem and then simplify. You can also simplify complex fractions by using the LCD of the fractions in the numerator and denominator.



Example 5A: Simplifying Complex Fractions

Simplify. Assume that all expressions are defined.

Write as division.

x + 2x – 1x – 3x + 5

Write the complex fraction as division.

x + 2x – 1

x – 3x + 5

÷

Multiply by the reciprocal.

x + 2x – 1

x + 5x – 3

Multiply.x2 + 7x + 10x2 – 4x + 3

(x + 2)(x + 5)(x – 1)(x – 3)

or




Write as division.

x + 1x2 – 1

x x – 1


x + 1x2 – 1

xx – 1

÷

Multiply by the reciprocal.x + 1x2 – 1

x – 1x




1x



x + 1(x – 1)(x + 1)

x – 1x


Factor the denominator.



Check It Out! Example 5b


Write as division.

20x – 1

63x – 3


20x – 1

63x – 3

÷

Multiply by the reciprocal.20x – 1

3x – 36



Check It Out! Example 5b Continued



20x – 1

3(x – 1)6


10



Check It Out! Example 5c


The LCD is (2x)(x – 2).

Multiply the numerator and denominator of the complex fraction by the LCD of the fractions in the numerator and denominator.

x + 4x – 2

12x

1x +

x + 4x – 2

(2x)(x – 2)

12x

1x + (2x)(x – 2)(2x)(x – 2)



Check It Out! Example 5c Continued


Simplify.3x – 6

(x + 4)(2x)

(2)(x – 2) + (x – 2)(x + 4)(2x)

3(x – 2)2x(x + 4)

or




Lesson Quiz: Part I

2.

3x2 – 2x + 7 (x – 2)(x + 1)

x ≠ –1, 2

x – 9 x – 4

x ≠ 4, –4


1. 2x + 1 x – 2

+ x – 3x + 1

x x + 4

– x + 36x2 – 16

3. Find the least common multiple of x2 – 6x + 5 and x2 + x – 2.

(x – 5)(x – 1)(x + 2)



Lesson Quiz: Part II

1 x

48 mi/h

5. Tyra averages 40 mi/h driving to the airport during rush hour and 60 mi/h on the return trip late at night. What is Tyra’s average speed for the entire trip?

4. Simplify . Assume that all expressions are defined.

x + 2x2 – 4

x x – 2

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Holt McDougal Algebra 2 8-3 Adding and Subtracting Rational Expressions Add and subtract rational expressions. Simplify complex fractions. Objectives