Lesson 11-3
Example 1 Find Excluded Values
State the excluded value of each expression.
2y
a.
3y +1
Exclude the values for which 3y + 1 = 0.
3y + 1 = 0 The denominator cannot be zero.
3y = -1 Subtract 1 from each side.
1
y=- Divide each side by 3.
3
1
Therefore, y cannot equal - .
3
-b
b.
b 2 -9
Exclude the values for which b 2 – 9 = 0.
b2 – 9 = 0 The denominator cannot be zero.
(b – 3)(b + 3) = 0 Factor.
b – 3 = 0 or b + 3 = 0 Zero Product Property
b=3 b = -3
Therefore, b cannot equal 3 or -3.
Real-World Example 2 Use Rational Expressions
GEOMETRY The area of the rectangle at the right
is 15a4b2c. Write an expression that gives the length 2
3a b
of the rectangle.
?
Understand You know the area and the width of the rectangle.
A
Plan Use the formula for length of a rectangle, = w. Substitute 15a4b2c for A and
3a2b for w.
A
Solve =w Write the equation.
15a4b2c 4 2 2
= Replace A with 15a b c and w with 3a b.
3a2b
821
= Simplify the denominator.
49
5.3 Simplify. Use a calculator.
So, the height of the cylinder is about 5.3 inches.
Check Use estimation to determine whether the answer is reasonable.
821
49(3)
5.6 or 5 The solution is reasonable.
Test Example 3 Expressions Involving Monomials
30x 2 y 3
Which expression is equivalent to ?
25xz
6xy 2 xy 3 63 6 xy 3
A B C D
z 5z 5 5z
Read the Test Item
30 x 2 y 3
The expression is a monomial divided by a monomial.
25 xz
Solve the Test Item
Step 1 The GCF of the numerator and denominator is 5x
5 x 6 xy 3
5 x 5 z
Step 2 Divide the numerator and denominator by the GCF.
5 x 6 xy 3
5 x 5 z
6 xy 3
Step 3 Simplify. The correct answer is D.
5z
Example 4 Simplify Rational Expressions
x2 – 2x – 8
Simplify . State the excluded values of x.
x+2
x2 – 2x – 8 (x + 2)(x – 4)
x+2 = x+2 Factor.
1
(x + 2)(x – 4) Divide the numerator and denominator by
= x+2 the GCF, x + 2.
1
=x–4 Simplify.
Exclude the values for which x + 2 equals 0.
x+2=0 The denominator cannot equal zero.
x = –2 Subtract 2 from each side.
So, x –2.
Example 5 Recognize Opposites
45 – 9x
Simplify 2 . State the excluded values of x.
x – 3x – 10
45 – 9x 9(5 – x)
= Factor.
x2 – 3x – 10 (x – 5)(x + 2)
9(–1)(x – 5)
= (x – 5)(x + 2) Rewrite 5 – x as –1(x – 5).
1
9(–1)(x – 5)
= Divide out the common factor, x – 5.
(x – 5)(x + 2)
1
9
=–x+2 Simplify.
Exclude the values for which x2 – 3x – 10 equals 0.
x2 – 3x – 10 = 0 The denominator cannot equal zero.
(x – 5)(x + 2) = 0 Factor.
x = 5 or x = –2 Zero Product Property
So, x –5 and x –2.
Example 6 Rational Functions
x2 – 8x – 7
Find the roots of f(x) = .
x+2
x2 – 8x – 7
f(x) = x+2 Original function
x2 – 8x – 7
0= x+2 f(x) = 0
(x – 1)(x – 7)
0= x+2 Factor.
When x = 1 or x = 7, the numerator becomes 0, so f(x) = 0. Therefore, the roots of the function are
1 and 7.