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```Adding
and
Subtracting
and
Subtracting
8-3
8-3 Rational Expressions
Rational Expressions
Warm Up
Lesson Presentation
Lesson Quiz
HoltMcDougal
Algebra 2Algebra 2
Holt
8-3
Rational Expressions
Warm Up
2 +
5
11 –
2. 12
1.
7
15
3
8
13
15
13
24
Simplify. Identify any x-values for which the
expression is undefined.
9
4x
3.
1 x6 x ≠ 0
3
12x
3
4.
x– 1
x2 – 1
Holt McDougal Algebra 2
1
x+1
x ≠ –1, x ≠ 1
8-3
Rational Expressions
Objectives
Simplify complex fractions.
Holt McDougal Algebra 2
8-3
Rational Expressions
Vocabulary
complex fraction
Holt McDougal Algebra 2
8-3
Rational Expressions
Adding and subtracting rational expressions is
or subtract rational expressions with like
denominators, add or subtract the numerators
and use the same denominator.
Holt McDougal Algebra 2
8-3
Rational Expressions
Example 1A: Adding and Subtracting Rational
Expressions with Like Denominators
Add or subtract. Identify any x-values for
which the expression is undefined.
x–3 + x–2
x+4
x+4
x–3+x–2
x+4
2x – 5
Combine like terms.
x+4
The expression is undefined at x = –4 because
this value makes x + 4 equal 0.
Holt McDougal Algebra 2
8-3
Rational Expressions
Example 1B: Adding and Subtracting Rational
Expressions with Like Denominators
Add or subtract. Identify any x-values for
which the expression is undefined.
3x – 4 – 6x + 1
x2 + 1
x2 + 1
3x – 4 – (6x + 1)
x2 + 1
3x – 4 – 6x – 1
x2 + 1
–3x – 5
x2 + 1
There is no real value of
the expression is always
Holt McDougal Algebra 2
Subtract the numerators.
Distribute the negative sign.
Combine like terms.
x for which x2 + 1 = 0;
defined.
8-3
Rational Expressions
Check It Out! Example 1a
Add or subtract. Identify any x-values for
which the expression is undefined.
6x + 5 + 3x – 1
x2 – 3
x2 – 3
6x + 5 + 3x – 1
x2 – 3
9x + 4
x2 – 3
Combine like terms.
The expression is undefined at x = ±
this value makes x2 – 3 equal 0.
Holt McDougal Algebra 2
because
8-3
Rational Expressions
Check It Out! Example 1b
Add or subtract. Identify any x-values for
which the expression is undefined.
3x2 – 5 – 2x2 – 3x – 2
3x – 1
3x – 1
3x2 – 5 – (2x2 – 3x – 2) Subtract the numerators.
3x – 1
3x2 – 5 – 2x2 + 3x + 2
Distribute the negative sign.
3x – 1
x2 + 3x – 3
Combine like terms.
3x – 1
1 because
The expression is undefined at x = 3
this value makes 3x – 1 equal 0.
Holt McDougal Algebra 2
8-3
Rational Expressions
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.
Holt McDougal Algebra 2
8-3
Rational Expressions
Example 2: Finding the Least Common Multiple of
Polynomials
Find the least common multiple for each pair.
A. 4x2y3 and 6x4y5
4x2y3 = 2  2  x2  y3
6x4y5 = 3  2  x4  y5
The LCM is 2  2  3  x4  y5, or 12x4y5.
B. x2 – 2x – 3 and x2 – x – 6
x2 – 2x – 3 = (x – 3)(x + 1)
x2 – x – 6 = (x – 3)(x + 2)
The LCM is (x – 3)(x + 1)(x + 2).
Holt McDougal Algebra 2
8-3
Rational Expressions
Check It Out! Example 2
Find the least common multiple for each pair.
a. 4x3y7 and 3x5y4
4x3y7 = 2  2  x3  y7
3x5y4 = 3  x5  y4
The LCM is 2  2  3  x5  y7, or 12x5y7.
b. x2 – 4 and x2 + 5x + 6
x2 – 4 = (x – 2)(x + 2)
x2 + 5x + 6 = (x + 2)(x + 3)
The LCM is (x – 2)(x + 2)(x + 3).
Holt McDougal Algebra 2
8-3
Rational Expressions
To add rational expressions with unlike
denominators, rewrite both expressions with
the LCD. This process is similar to adding
fractions.
Holt McDougal Algebra 2
8-3
Rational Expressions
Add. Identify any x-values for which the
expression is undefined.
x–3
2x
+
x2 + 3x – 4
x+4
x–3
2x
+
(x + 4)(x – 1) x + 4
x–3
2x
+
(x + 4)(x – 1) x + 4
Holt McDougal Algebra 2
Factor the denominators.
+ 4)(x – 1),
x – 1 The LCD is (x2x
x– 1.
so
multiply
by
x–1
x–1
x+4
8-3
Rational Expressions
Example 3A Continued
Add. Identify any x-values for which the
expression is undefined.
x – 3 + 2x(x – 1)
(x + 4)(x – 1)
2x2 – x – 3
(x + 4)(x – 1)
Simplify the numerator.
Write the sum in
factored or expanded
form.
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.
2x2 – x – 3 or
(x + 4)(x – 1)
Holt McDougal Algebra 2
2x2 – x – 3
x2 + 3x – 4
8-3
Rational Expressions
Add. Identify any x-values for which the
expression is undefined.
x
+ 2–8
x+2
x –4
x +
–8
x + 2 (x + 2)(x – 2)
Factor the denominator.
x
–8
x–2 +
The LCD is (x + 2)(x – 2),
x
x+ 2 x – 2
(x + 2)(x – 2) so multiply
by x – 2 .
x–2
x+2
x(x – 2) + (–8)
(x + 2)(x – 2)
Holt McDougal Algebra 2
8-3
Rational Expressions
Example 3B Continued
Add. Identify any x-values for which the
expression is undefined.
x2 – 2x – 8
(x + 2)(x – 2)
Write the numerator in
standard form.
(x + 2)(x – 4)
(x + 2)(x – 2)
Factor the numerator.
x–4
x–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.
Holt McDougal Algebra 2
8-3
Rational Expressions
Check It Out! Example 3a
Add. Identify any x-values for which the
expression is undefined.
3x + 3x – 2
2x – 2 3x – 3
3x + 3x – 2
2(x – 1) 3(x – 1)
3x
3 + 3x – 2 2
2(x – 1) 3
3(x – 1) 2
Holt McDougal Algebra 2
Factor the denominators.
The LCD is 6(x – 1), so
multiply 3x
by 3 and
2(x – 1)
3x – 2 by 2.
3(x – 1)
8-3
Rational Expressions
Check It Out! Example 3a Continued
Add. Identify any x-values for which the
expression is undefined.
9x + 6x – 4
6(x – 1)
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.
Holt McDougal Algebra 2
8-3
Rational Expressions
Check It Out! Example 3b
Add. Identify any x-values for which the
expression is undefined.
x
x+3
+
2x + 6
x2 + 6x + 9
x
x+3
2x + 6
+
(x + 3)(x + 3)
Factor the denominators.
x
x+3+
2x + 6
The LCD is (x + 3)(x + 3),
x + 3 x + 3 (x + 3)(x + 3)
x
(x + 3)
so multiply (x + 3) by (x + 3) .
x2 + 3x + 2x + 6
(x + 3)(x + 3)
Holt McDougal Algebra 2
8-3
Rational Expressions
Check It Out! Example 3b Continued
Add. Identify any x-values for which the
expression is undefined.
x2 + 5x + 6
(x + 3)(x + 3)
(x + 3)(x + 2)
(x + 3)(x + 3)
x+2
x+3
Write the numerator in
standard form.
Factor the numerator.
Divide out common
factors.
The expression is undefined at x = –3 because this
value of x make the factors (x + 3) and (x + 3)
equal 0.
Holt McDougal Algebra 2
8-3
Rational Expressions
Example 4: Subtracting Rational Expressions
2 – 30
2x
x + 5 . Identify any xSubtract
–
2
x –9
x+3
values for which the expression is undefined.
2x2 – 30
x+5
–
(x – 3)(x + 3)
x+3
Factor the denominators.
2x2 – 30
x + 5 x – 3 The LCD is (x – 3)(x + 3),
–
x+5
(x – 3)
(x – 3)(x + 3)
x + 3 x – 3 so multiply
x + 3 by (x – 3) .
2x2 – 30 – (x + 5)(x – 3)
(x – 3)(x + 3)
Subtract the numerators.
2x2 – 30 – (x2 + 2x – 15)
(x – 3)(x + 3)
Multiply the binomials in
the numerator.
Holt McDougal Algebra 2
8-3
Rational Expressions
Example 4 Continued
2 – 30
2x
x + 5 . Identify any xSubtract
–
x2 – 9
x+3
values for which the expression is undefined.
2x2 – 30 – x2 – 2x + 15
Distribute the negative sign.
(x – 3)(x + 3)
x2 – 2x – 15
Write the numerator in
(x – 3)(x + 3)
standard form.
(x + 3)(x – 5)
Factor the numerator.
(x – 3)(x + 3)
x–5
Divide out common factors.
x–3
The expression is undefined at x = 3 and x = –3
because these values of x make the factors (x + 3)
and (x – 3) equal 0.
Holt McDougal Algebra 2
8-3
Rational Expressions
Check It Out! Example 4a
3x – 2
2
. Identify any x–
2x + 5
5x – 2
values for which the expression is undefined.
Subtract
3x – 2 5x – 2
2
2x + 5 The LCD is (2x + 5)(5x – 2),
–
2x + 5 5x – 2 5x – 2 2x + 5 so multiply 3x – 2 by (5x – 2)
2x + 5
(5x – 2)
2 by (2x + 5) .
and
5x – 2
(2x + 5)
(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.
Holt McDougal Algebra 2
8-3
Rational Expressions
Check It Out! Example 4a Continued
3x – 2
2
. Identify any x–
2x + 5
5x – 2
values for which the expression is undefined.
Subtract
15x2 – 16x + 4 – 4x – 10
(2x + 5)(5x – 2)
Distribute the negative sign.
The expression is undefined at x = – 5 and x = 2
2
5
because these values of x make the factors (2x + 5)
and (5x – 2) equal 0.
Holt McDougal Algebra 2
8-3
Rational Expressions
Check It Out! Example 4b
2 + 64
2x
x – 4 . Identify any xSubtract
–
2
x – 64
x+8
values for which the expression is undefined.
2x2 + 64
x–4
–
(x – 8)(x + 8)
x+8
Factor the denominators.
The LCD is (x – 3)(x + 8),
2x2 + 64
x–4 x–8
x–4
(x – 8)
–
(x – 8)(x + 8)
x + 8 x – 8 so multiply x + 8 by (x – 8) .
2x2 + 64 – (x – 4)(x – 8)
(x – 8)(x + 8)
Subtract the numerators.
2x2 + 64 – (x2 – 12x + 32)
(x – 8)(x + 8)
Multiply the binomials in
the numerator.
Holt McDougal Algebra 2
8-3
Rational Expressions
Check It Out! Example 4b
2 + 64
2x
x – 4 . Identify any xSubtract
–
x2 – 64
x+8
values for which the expression is undefined.
2x2 + 64 – x2 + 12x – 32)
Distribute the negative sign.
(x – 8)(x + 8)
x2 + 12x + 32
Write the numerator in
(x – 8)(x + 8)
standard form.
(x + 8)(x + 4)
Factor the numerator.
(x – 8)(x + 8)
x+4
Divide out common factors.
x–8
The expression is undefined at x = 8 and x = –8
because these values of x make the factors (x + 8)
and (x – 8) equal 0.
Holt McDougal Algebra 2
8-3
Rational Expressions
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.
Holt McDougal Algebra 2
8-3
Rational Expressions
Example 5A: Simplifying Complex Fractions
Simplify. Assume that all expressions are defined.
x+2
x–1
x–3
x+5
Write the complex fraction as division.
x+2 ÷ x–3
Write as division.
x–1
x+5
Multiply by the
x+2
x+5

reciprocal.
x–1
x–3
(x + 2)(x + 5) or x2 + 7x + 10
(x – 1)(x – 3)
x2 – 4x + 3
Holt McDougal Algebra 2
Multiply.
8-3
Rational Expressions
Example 5B: Simplifying Complex Fractions
Simplify. Assume that all expressions are defined.
3
x
+ 2
x
x–1
x
Multiply the numerator and denominator of the
complex fraction by the LCD of the fractions in
the numerator and denominator.
3
x
(2x) +
(2x)
x
2
x – 1 (2x)
x
Holt McDougal Algebra 2
The LCD is 2x.
8-3
Rational Expressions
Example 5B Continued
Simplify. Assume that all expressions are defined.
(3)(2) + (x)(x)
(x – 1)(2)
Divide out common
factors.
x2 + 6 or
2(x – 1)
Simplify.
Holt McDougal Algebra 2
x2 + 6
2x – 2
8-3
Rational Expressions
Check It Out! Example 5a
Simplify. Assume that all expressions are defined.
x+1
x2 – 1
x
x–1
Write the complex fraction as division.
x+1 ÷
x
x2 – 1
x–1
Write as division.
x+1 
x2 – 1
Multiply by the reciprocal.
x–1
x
Holt McDougal Algebra 2
8-3
Rational Expressions
Check It Out! Example 5a Continued
Simplify. Assume that all expressions are defined.
x+1
 x–1
(x – 1)(x + 1)
x
1
x
Holt McDougal Algebra 2
Factor the denominator.
Divide out common factors.
8-3
Rational Expressions
Check It Out! Example 5b
Simplify. Assume that all expressions are defined.
20
x–1
6
3x – 3
Write the complex fraction as division.
20 ÷
6
x–1
3x – 3
Write as division.
20  3x – 3
x–1
6
Multiply by the reciprocal.
Holt McDougal Algebra 2
8-3
Rational Expressions
Check It Out! Example 5b Continued
Simplify. Assume that all expressions are defined.
20  3(x – 1)
x–1
6
10
Holt McDougal Algebra 2
Factor the numerator.
Divide out common factors.
8-3
Rational Expressions
Check It Out! Example 5c
Simplify. Assume that all expressions are defined.
1
1
+ 2x
x
x+4
x–2
Multiply the numerator and denominator of the
complex fraction by the LCD of the fractions in
the numerator and denominator.
1
1
(2x)(x – 2) +
(2x)(x – 2)
x
2x
x + 4 (2x)(x – 2)
x–2
Holt McDougal Algebra 2
The LCD is (2x)(x – 2).
8-3
Rational Expressions
Check It Out! Example 5c Continued
Simplify. Assume that all expressions are defined.
(2)(x – 2) + (x – 2)
(x + 4)(2x)
3x – 6 or 3(x – 2)
(x + 4)(2x) 2x(x + 4)
Holt McDougal Algebra 2
Divide out common
factors.
Simplify.
8-3
Rational Expressions
Example 6: Transportation Application
A hiker averages 1.4 mi/h when walking downhill
on a mountain trail and 0.8 mi/h on the return trip
when walking uphill. What is the hiker’s average
speed for the entire trip? Round to the nearest
tenth.
Total distance: 2d
Let d represent the
one-way distance.
Total time: d + d
1.4 0.8
Use the formula t = d
r .
Average speed:
Holt McDougal Algebra 2
2d
d + d
1.4 0.8
The average speed is
total distance .
total time
8-3
Rational Expressions
Example 6 Continued
2d
d = d = 5d and d = d = 5d .
1.4
7
7
0.8
4
4
5
5
2d(28)
The LCD of the fractions in the
5d (28) + 5d (28)
denominator is 28.
7
4
56d
Simplify.
20d + 35d
d + d
1.4 0.8
55d
≈ 1.0
55d
Combine like terms and
divide out common factors.
The hiker’s average speed is 1.0 mi/h.
Holt McDougal Algebra 2
8-3
Rational Expressions
Check It Out! Example 6
Justin’s average speed on his way to school is 40
mi/h, and his average speed on the way home is
45 mi/h. What is Justin’s average speed for the
entire trip? Round to the nearest tenth.
Total distance: 2d
Let d represent the
one-way distance.
Total time: d + d
40
45
Use the formula t = d
r .
Average speed:
Holt McDougal Algebra 2
2d
d + d
40
45
The average speed is
total distance .
total time
8-3
Rational Expressions
Check It Out! Example 6
2d(360)
d (360)+ d (360)
40
45
720d
9d + 8d
720d ≈ 42.4
17d
The LCD of the fractions in the
denominator is 360.
Simplify.
Combine like terms and
divide out common factors.
Justin’s average speed is 42.4 mi/h.
Holt McDougal Algebra 2
8-3
Rational Expressions
Lesson Quiz: Part I
Add or subtract. Identify any x-values for
which the expression is undefined.
1. 2x + 1 + x – 3
x–2
x+1
2.
x
x+4
– x2 + 36
x – 16
3x2 – 2x + 7 x ≠ –1, 2
(x – 2)(x + 1)
x–9
x–4
x ≠ 4, –4
3. Find the least common multiple of x2 – 6x + 5
and x2 + x – 2.
(x – 5)(x – 1)(x + 2)
Holt McDougal Algebra 2
8-3
Rational Expressions
Lesson Quiz: Part II
4. Simplify
defined.
x+2
x2 – 4
x
x–2
. Assume that all expressions are
1
x
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? 48 mi/h
Holt McDougal Algebra 2
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