Intermediate Algebra Unit 6 Rational Expressions

Intermediate Algebra Exam 2 Material Rational Expressions

Rational Expression • A ratio of two polynomials where the denominator is not zero (an “ugly fraction” with a variable in a denominator) • Example: x2 − x − 2 x+3 • Will the value of the denominator ever be zero? If x = - 3, then the denominator becomes 0, so we say that – 3 is a restricted value of x • What is the “domain” of the rational expression (all acceptable values of the variable)? Domain is the set of all real numbers except - 3. Domain: {x | x ≠ -3}

Finding Restricted Values and Domains of Rational Expressions • Completely factor the denominator • Make equations by setting each factor of the denominator equal to zero • Solve the equations to find restricted values • The domain will be all real numbers that are not restricted

Example Find the domain:

3x − 5 x2 − 4

• Factor the denominator: (x – 2)(x + 2) • Set each factor equal to zero and solve the equations: x – 2 = 0 and x + 2 = 0 x=2 and x = -2 (Restricted Values) • Domain: {x | x ≠ -2, x ≠ 2}

Evaluating Rational Expressions • To “evaluate” a rational expression means to find its “value” when variables are replaced by specific “unrestricted” numbers inside parentheses 2x − 3 • Example: Evaluate 2 for x = -2 2(− 2) − 3 = 2 (− 2) + 2( − 2) − 3

x + 2x − 3

−4−3 = 4−4−3

−7 = −3

7 3

Fundamental Principle of Fractions • If the numerator and denominator of a fraction contain a common factor, that factor may be divided out to reduce the fraction to lowest terms: 1

ab b = c 1 ac

1

1

12 2⋅ 2⋅3 2 = = 18 12 ⋅13 ⋅ 3 3

When common factors are divided out, "1" is left in each place. 1

1 5 = 15 x x

Reducing Rational Expressions to Lowest Terms • Completely factor both numerator and denominator • Apply the fundamental principle of fractions: divide out common factors that are found in both the numerator and the denominator

Example of Reducing Rational Expressions to Lowest Terms • Reduce to lowest terms:

3 x 3 + 24 3x + 6

(

• Factor top and bottom: 1

)

3 x3 + 8 3( x + 2 )

(

1

3( x + 2 ) x 2 − 2 x + 4 3( x + 2 ) 1

• Divide out common factors to get:

1

x2 − 2x + 4

)

Example of Reducing Rational Expressions to Lowest Terms • Reduce to lowest terms: • Factor top and bottom:

x −3 3− x x −3 1( x − 3) = − x+3 − 1( x − 3) 1( x − 3) = − 1( x − 3)

• Divide out common factors to get:

−1

1

Equivalent Forms of Rational Expressions • All of the following are equivalent: −p p p = =− q −q q

• In words this would say that a negative factor in the numerator or denominator can be moved, respectively, to a negative factor in the denominator or numerator, or can be moved to the front of the fraction, or vice versa

Example of Using Equivalent Forms of Rational Expressions • Write equivalent forms of:

x−5 − ( x − 5) − = = x+2 x+2

− x+5 x+2

x −5 x−5 = − = − ( x + 2) x+2

x −5 −x−2

Homework Problems • Section: 6.1 • Page: 401 • Problems: Odd: 3 – 9, 13 – 23, 27 – 63, 67 – 73 • MyMathLab Homework Assignment 6.1 for practice • MyMathLab Quiz 6.1 for grade

Multiplying Rational Expressions (Same as Multiplying Fractions) • Factor each numerator and denominator • Divide out common factors • Write answer (leave polynomials in factored form) • Example: 1 1 1 5 2 ⋅ 2 3⋅5 4 15 ⋅ = ⋅ = 21 3⋅3 2 ⋅ 2 ⋅ 7 9 28 1 1 1

Example of Multiplying Rational Expressions 3x 2 − 2 x − 8 3x + 2 ⋅ 2 3 x + 14 x + 8 3 x + 4

Completely factor each top and bottom: 1

1

( 3x + 4)( x − 2) ⋅ ( 3x + 2) ( 3x1+ 2)( x + 4) ( 3x + 4) 1

Divide out common factors:

( x − 2) ( x + 4)

Dividing Rational Expressions (Same as Dividing Fractions) • Invert the divisor and change problem to multiplication a c a d ad ÷ = ⋅ = b d b c bc

• Example: 2 3 2 4 ÷ = ⋅ = 3 4 3 3

8 9

Example of Dividing Rational Expressions 2y 8y + 4y 2y 27 ÷ = ⋅ 5 = 3 9 27 9 8y + 4y 2

5

3

2

3 2 y 2 y2 +1

(

)

1 1

3

2

2y 27 ⋅ 3 2 = 9 4 y 2 y +1 1

2 y

(

)

Homework Problems • Section: 6.2 • Page: 408 • Problems: Odd: 3 – 25, 29 – 61 • MyMathLab Homework Assignment 6.2 for practice • MyMathLab Quiz 6.2 for grade

Finding the Least Common Denominator, LCD, of Rational Expressions • Completely factor each denominator • Construct the LCD by writing down each factor the maximum number of times it is found in any denominator

Example of Finding the LCD • Given three denominators, find the LCD: 2 2 6 x − 12 , , 4 x − 16 x + 16 3 x − 12 • Factor each denominator:

3 x 2 −12 = 3( x 2 − 4 ) = 3( x − 2 )( x + 2 ) 6x −12 = 6( x − 2 ) = 2 ⋅ 3( x − 2) 2 4 x 2 − 16 x + 16 = 4( x − 4 x + 4 ) = 2 ⋅ 2( x − 2 )( x − 2 )

• Construct LCD by writing each factor the maximum number of times it’s found in any denominator:

LCD = 2 ⋅ 2 ⋅ 3( x − 2 )( x − 2 )( x + 2 ) LCD = 12( x − 2 ) 2 ( x + 2 )

Equivalent Fractions • The fundamental principle of fractions, mentioned earlier, says:

ab b = ac c

• In words, this says that when numerator and denominator of a fraction are multiplied by the same factor, the result is equivalent to the original fraction

2 6⋅2 = = 3 6⋅3

12 18

.

Writing Equivalent Fractions With Specified Denominator • Given a fraction and a desired denominator for an equivalent fraction that is a multiple of the original denominator, write an equivalent fraction by multiplying both the numerator and denominator of the original fraction by all factors of the desired denominator not found in the original denominator • To accomplish this goal, it is usually best to completely factor both the original denominator and the desired denominator

Example Write an equivalent fraction to the given fraction that has a denominator of 24:

5 ? = 6 24 5 2⋅ 2⋅5 20 = = 6 2⋅2⋅6 24

Factor each denominator :

6 = 2⋅3 24 = 2 ⋅ 2 ⋅ 2 ⋅ 3

Example Write an equivalent rational expression to the given one that has a denominator of Factor each denominator : 2 y3 − 4 y2 + 2 y : y+2 ? = 3 2 y − y 2 y − 4 y2 + 2 y

y 2 − y = y ( y − 1)

2 y 3 − 4 y 2 + 2 y = 2 y ( y − 1)( y − 1)

(

)

y+2 y+2 ( y + 2 ) 2( y − 1) 2 y2 + y − 2 = = = = 2 y − y y ( y − 1) y ( y − 1) 2( y − 1) y ( y − 1) 2( y − 1) 2 y2 + 2 y − 4 2 y3 − 4 y2 + 2 y

Homework Problems • Section: 6.3 • Page: 414 • Problems: Odd: 5 – 43, 51 – 69 • MyMathLab Homework Assignment 6.3 for practice • MyMathLab Quiz 6.3 for grade

Adding and Subtracting Rational Expressions (Same as Fractions) • Find a least common denominator, LCD, for the rational expressions • Write each fraction as an equivalent fraction having the LCD • Write the answer by adding or subtracting numerators as indicated, and keeping the LCD • If possible, reduce the answer to lowest terms

Example y+2 3y 1 − 2 + 2 y − y 2y − 4y + 2 y y ( y − 1) 2( y − 1)( y − 1) y

=

y+2 3y 1 − + y ( y − 1) 2( y − 1)( y − 1) y

•

Find a least common denominator, LCD, for the rational expressions:

•

Write each fraction as an equivalent fraction having the LCD:

•

Write the answer by adding or subtracting numerators as indicated, and keeping the LCD:2 2 2 2 2 2

LCD = 2 y ( y − 1)( y − 1)

2( y + 2 )( y − 1) 3y ⋅ y 2 ⋅1( y − 1)( y − 1) − + 2 y ( y − 1)( y − 1) 2 y ( y − 1)( y − 1) 2 y ( y − 1)( y − 1)

(

)

(

)

2 y + 2 y − 4 − 3y + 2 y − 4 y + 2 2 y + y − 2 − 3y + 2 y − 2 y +1 = = = ( )( ) 2 y y − 1 y − 1 2 y ( y − 1)( y − 1) •

If possible, reduce the answer to lowest terms

y2 − 2 y − 2 2 y ( y − 1)( y − 1)

Since top won' t factor, fraction won' t reduce!

Homework Problems • Section: 6.4 • Page: 422 • Problems: Odd: 9 – 21, 25 – 47, 51 – 71 • MyMathLab Homework Assignment 6.4 for practice • MyMathLab Quiz 6.4 for grade

Complex Fraction • A “fraction” that contains a rational expression in its numerator, or in its denominator, or both • Example: 1 +2

3x 5 6y

• Think of it as “fractions inside of a fraction” • Every complex fraction can be simplified to a rational expression (ratio of two polynomials)

Two Methods for Simplifying Complex Fractions • Method One – Do math on top to get a single fraction – Do math on bottom to get a single fraction – Divide top fraction by bottom fraction

• Method Two (Usually preferred) – Find the LCD of all of the “little fractions” – Multiply the complex fraction by “1” where “1” is the LCD of the little fractions over itself

Method One Example of Simplifying a Complex Fraction 1 +2 3x 5 6y

•

Do math on top to get single fraction:

1 1 2 1 6x 1+ 6 x +2= + = + = 3x 3x 1 3x 3x 3x

•

Do math on bottom to get single fraction:

•

Top fraction divided by bottom:

5 In this case, bottom is already single fraction : 6y

2 1+ 6x 5 1+ 6x 6 y ÷ = ⋅ = 3x 6y 3x 5

2 y + 12 xy 5x

Method Two Example of Simplifying a Complex Fraction 1 +2 3x 5 6y • Find the LCD of all of the “little fractions”: 6 xy • Multiply the complex fraction by “1” where “1” is the LCD of the little fractions over itself  1  6 xy + 2   3x  1 = ⋅ = 6 xy  5    1  6y 

6 xy 12 xy + 3x 1 = 30 xy 6y

2 y + 12 xy 5x

Homework Problems • Section: 6.5 • Page: 431 • Problems: Odd: 7 – 35 • MyMathLab Homework Assignment 6.5 for practice • MyMathLab Quiz 6.5 for grade

Other Types of Equations • Thus far techniques have been discussed for solving all linear and some quadratic equations • Now address techniques for identifying and solving “rational equations”

Rational Equations • Technical Definition: An equation that contains a rational expression • Practical Definition: An equation that has a variable in a denominator • Example:

1 5 2 = − 2 x + 2x − 3 x −1 x + 3

Solving Rational Equations • Find “restricted values” for the equation by setting every denominator that contains a variable equal to zero and solving • Find the LCD of all the fractions and multiply both sides of equation by the LCD to eliminate fractions • Solve the resulting equation to find apparent solutions • Solutions are all apparent solutions that are not restricted

Example

RV

1 5 2 x + 2x − 3 = 0 = − 2 ( x − 1)( x + 3) = 0 x + 2x − 3 x −1 x + 3 x − 1 = 0 OR x + 3 = 0 x = −3 1 5 2 x =1 = − x − 1 = 0 Already Solved ( x − 1)( x + 3) x − 1 x + 3 x + 3 = 0 Already Solved  LCD 1 5 2  LCD ( x − 1)( x + 3)  ( x − 1)( x + 3) = x − 1 − x + 3  1   2

1 = 5( x + 3) − 2( x − 1) 1 = 5 x + 15 − 2 x + 2 1 = 3 x + 17

− 16 = 3 x 16 x=− 3

Not RV!

Example

RV

2 1 1 m −1 = 0 − = 2 ( m − 1)( m + 1) = 0 m −1 2 m −1 m − 1 = 0 OR m + 1 = 0 m = −1 2 1 1 m =1 − = m − 1 = 0 Already Solved ( m − 1)( m + 1) 2 m − 1  2 1 1  LCD LCD − =  ( m − 1)( m + 1) 2 m − 1 1 2( m − 1)( m + 1)   4 − ( m − 1)( m + 1) = 2( m + 1) 0 = ( m − 3)( m + 1) 4 − ( m 2 − 1) = 2m + 2 m − 3 = 0 or m + 1 = 0 2 4 − m + 1 = 2m + 2 m = 3 or m = −1 2 0 = m + 2m − 3 2

Formula • Any equation containing more than one variable • To solve a formula for a specific variable we must use appropriate techniques to isolate that variable on one side of the equal sign • The technique we use in solving depends on the type of equation for the variable for which we are solving

Example of Different Types of Equations for the Same Formula 2

4 B • Consider the formula: A + 3 = C −1

• What type of equation for A? Linear (variable to first power) • What type of equation for B? Quadratic (variable to second power) • What type of equation for C? Rational (variable in denominator)

Solving Formulas Involving Rational Equations •

Use the steps previously discussed for solving rational equations: •

• • •

Find “restricted values” for the equation by setting every denominator that contains the variable being solved for equal to zero and solving Find the LCD of all the fractions and multiply both sides of equation by the LCD to eliminate fractions Solve the resulting equation to find apparent solutions Solutions are all apparent solutions that are not restricted

Solve the Formula for C:

4B 2 A+3 = C −1 Since the formula is rational for C, find RV: C =1 C −1 = 0 Multiply both sides by LCD: 2 4B ( C −1)( A + 3) = ( C −1) ( C −1)

AC + 3C − A − 3 = 4 B 2

( C − 1)

Example Continued Solve resulting equation and check apparent answer with RV: 2 ( Now linear for C ) AC + 3C − A − 3 = 4 B AC + 3C = 4 B 2 + A + 3 2 ( A + 3) C = 4 B + A + 3 ( A + 3) C = 4 B 2 + A + 3 ( A + 3) ( A + 3) 4B 2 + A + 3 C= Not RV A+3

Homework Problems • Section: 6.6 • Page: 439 • Problems: Odd: 17 – 69, 73 – 87 • MyMathLab Homework Assignment 6.6 for practice • ( No MyMathLab Quiz until we finish Section 6.7 )

Applications of Rational Expressions • Word problems that translate to rational expressions are handled the same as all other word problems • On the next slide we give an example of such a problem

Example When three more than a number is divided by twice the number, the result is the same as the original number. Find all numbers that satisfy these conditions. RV : Unknowns : The number x x+3 =x 2x  x +3 2 x  = 2 x( x )  2x  x + 3 = 2x2

0 = 2x2 − x − 3

0 = ( 2 x − 3)( x + 1)

2x = 0 x=0

2 x − 3 = 0 or x + 1 = 0 2 x = 3 or x = −1 3 x = or x = −1 2

.

Homework Problems • Section: 6.7 • Page: 449 • Problems: Odd: 3 –9 • MyMathLab Homework Assignment 6.7 for practice • MyMathLab Quiz 6.6 - 6.7 for grade