Transcript Intermediate Algebra, 5ed

Chapter 6
Rational
Expressions
§ 6.1
Rational Functions and
Multiplying and Dividing
Rational Expressions
Rational Expressions
Rational expressions can be written in the form
where P and Q are both polynomials and Q  0.
P
Q
Examples of Rational Expressions
3x 2  2 x  4
4x  5
4x  3y
2
2
2 x  3 xy  4 y
Martin-Gay, Intermediate Algebra, 5ed
3x 2
4
3
Evaluating Rational Expressions
To evaluate a rational expression for a particular
value(s), substitute the replacement value(s) into the
rational expression and simplify the result.
Example:
Evaluate the following expression for y = 2.
y2
4 4
2  2



 5  y 5  (2)  7 7
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Evaluating Rational Expressions
In the previous example, what would happen if we
tried to evaluate the rational expression for y = 5?
y2
3
5

2


 5  y 5  5 0
This expression is undefined!
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Undefined Rational Expressions
We have to be able to determine when a
rational expression is undefined.
A rational expression is undefined when the
denominator is equal to zero.
The numerator being equal to zero is okay
(the rational expression simply equals zero).
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Undefined Rational Expressions
Find any real numbers that make the following rational
expression undefined.
Example:
9x3  4 x
15x  45
The expression is undefined when 15x + 45 = 0.
So the expression is undefined when x = 3.
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Simplifying Rational Expressions
Simplifying a rational expression means writing it in
lowest terms or simplest form.
Fundamental Principle of Rational Expressions
For any rational expression P and any polynomial R,
Q
where R ≠ 0,
PR P R P
P
   1 
QR Q R Q
Q
PR P
 .
or, simply,
QR Q
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Simplifying Rational Expressions
Simplifying a Rational Expression
1) Completely factor the numerator and
denominator of the rational expression.
2) Divide out factors common to the numerator and
denominator. (This is the same thing as
“removing the factor of 1.”)
Warning!
Only common FACTORS can be eliminated from the
numerator and denominator. Make sure any
expression you eliminate is a factor.
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Simplifying Rational Expressions
Example:
Simplify the following expression.
7( x  5) 7
7 x  35


2
x( x  5)
x  5x
x
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Simplifying Rational Expressions
Example:
Simplify the following expression.
( x  4)(x  1)
x  3x  4
x 1


2
x  x  20 ( x  5)(x  4)
x5
2
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Simplifying Rational Expressions
Example:
Simplify the following expression.
7  y  1( y  7)
 1

y 7
y 7
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Multiplying Rational Expressions
Multiplying Rational Expressions
The rule for multiplying rational expressions is
P R PR
as long as Q  0 and S  0.
 
Q S QS
1) Completely factor each numerator and denominator.
2) Use the rule above and multiply the numerators and
denominators.
3) Simplify the product by dividing the numerator and
denominator by their common factors.
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Multiplying Rational Expressions
Example:
Multiply the following rational expressions.
2 3 x  x 5 x
6 x 5x
1



3
10x 12 2  5  x  x  x  2  2  3
4
2
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Multiplying Rational Expressions
Example:
Multiply the following rational expressions.
( m  n)
m
(m  n)(m  n)  m
 2


m  n m  m n (m  n)  m(m  n)
2
mn
mn
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Dividing Rational Expressions
Dividing Rational Expressions
The rule for dividing rational expressions is
P R P S PS
   
Q S Q R QR
as long as Q  0 and S  0 and R  0.
To divide by a rational expression, use the rule
above and multiply by its reciprocal. Then simplify
if possible.
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Dividing Rational Expressions
Example:
Divide the following rational expression.
25
( x  3) 5 x  15 ( x  3)




5
5 x  15
5
25
2
2
( x  3)(x  3)  5  5
 x3
5  5( x  3)
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§ 6.2
Adding and Subtracting
Rational Expressions
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Adding and Subtracting with Common
Denominators
Adding or Subtracting Rational Expressions
with Common Denominators
The rule for multiplying rational expressions is
P
R are rational expression, then
If
and
Q
Q
P R PR
P R PR
 
and  
Q Q
Q
Q Q
Q
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Adding Rational Expressions
Example:
Add the following rational expressions.
4p 3 3p 8 7 p  5
4 p 3 3p 8



2p 7
2p 7
2p 7 2p 7
Martin-Gay, Intermediate Algebra, 5ed
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Subtracting Rational Expressions
Example:
Subtract the following rational expressions.
8 y  16 8( y  2)
8y
16



 8
y2
y2 y2
y2
Martin-Gay, Intermediate Algebra, 5ed
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Subtracting Rational Expressions
Example:
Subtract the following rational expressions.
3y  6
3y
6

 2
 2
2
y  3 y  10
y  3 y  10 y  3 y  10
3( y  2)

( y  5)( y  2)
3
y5
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Least Common Denominators
To add or subtract rational expressions with
unlike denominators, you have to change
them to equivalent forms that have the same
denominator (a common denominator).
This involves finding the least common
denominator of the two original rational
expressions.
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Least Common Denominators
Finding the Least Common Denominator
1) Factor each denominator completely,
2) The LCD is the product of all unique factors each
raised to a power equal to the greatest number of
times that the factor appears in any factored
denominator.
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Least Common Denominators
Example:
Find the LCD of the following rational expressions.
1
3x
,
6 y 4 y  12
6 y  2  3y
4 y  12  4( y  3)  2 ( y  3)
2
So theLCD is 2  3 y( y  3)  12y( y  3)
2
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Least Common Denominators
Example:
Find the LCD of the following rational expressions.
4
4x  2
, 2
2
x  4 x  3 x  10 x  21
x  4x  3  ( x  3)(x  1)
2
x  10x  21  ( x  3)(x  7)
2
So theLCD is (x  3)(x 1)(x 7)
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Least Common Denominators
Example:
Find the LCD of the following rational expressions.
2
3x
4x
, 2
2
5x  5 x  2 x  1
5x  5  5( x 1)  5( x  1)(x 1)
2
2
x  2x  1  ( x 1)
2
2
So theLCD is 5(x  1)(x-1)2
Martin-Gay, Intermediate Algebra, 5ed
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Least Common Denominators
Example:
Find the LCD of the following rational expressions.
1
2
,
x 3 3 x
Both of the denominators are already factored.
Since each is the opposite of the other, you can
use either x – 3 or 3 – x as the LCD.
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Multiplying by 1
To change rational expressions into equivalent
forms, we use the principal that multiplying
by 1 (or any form of 1), will give you an
equivalent expression.
P P
P R PR
 1   
Q Q
Q R QR
Martin-Gay, Intermediate Algebra, 5ed
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Equivalent Expressions
Example:
Rewrite the rational expression as an equivalent
rational expression with the given denominator.
3

5
9y
72 y 9
4
3 8y
24y 4
3
 4 

5
5
9
9y 8y
9y
72y
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Unlike Denominators
As stated in the previous section, to add or
subtract rational expressions with different
denominators, we have to change them to
equivalent forms first.
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Unlike Denominators
Adding or Subtracting Rational Expressions
with Unlike Denominators
1) Find the LCD of the rational expressions.
2) Write each rational expression as an equivalent
rational expression whose denominator is the LCD
found in Step 1.
3) Add or subtract numerators, and write the result
over the denominator.
4) Simplify resulting rational expression, if possible.
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Adding with Unlike Denominators
Example:
Add the following rational expressions.
15 8
,
7 a 6a
15 8
6 15 7  8




7 a 6 a 6  7 a 7  6a
73
90
56
146



21a
42 a 42 a 42 a
Martin-Gay, Intermediate Algebra, 5ed
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Subtracting with Unlike Denominators
Example:
Subtract the following rational expressions.
5
3
,
2x  6 6  2x
5
3
5
3




2x  6 6  2x 2x  6 2x  6
4
222
8


2 x  6 2( x  3) x  3
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Subtracting with Unlike Denominators
Example:
Subtract the following rational expressions.
7
and 3
2x  3
7
3(2 x  3)
7


3 
2x  3
2x  3
2x  3
7
6 x  9 7  6 x  9 16  6 x



2x  3
2x  3 2x  3
2x  3
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Adding with Unlike Denominators
Example:
Add the following rational expressions.
4
x
, 2
2
x  x  6 x  5x  6
4
x
4
x
 2



2
x  x  6 x  5 x  6 ( x  3)(x  2) ( x  3)(x  2)
4( x  3)
x( x  3)


( x  3)(x  2)(x  3) ( x  3)(x  2)(x  3)
2
2
x
 x  12
4 x  12  x  3x

( x  2)(x  3)(x  3)
( x  2)(x  3)(x  3)
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§ 6.3
Simplifying Complex
Fractions
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Complex Rational Fractions
Complex rational expressions (complex
fraction) are rational expressions whose
numerator, denominator, or both contain one or
more rational expressions.
There are two methods that can be used when
simplifying complex fractions.
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Simplifying Complex Fractions
Simplifying a Complex Fraction (Method 1)
1) Simplify the numerator and the denominator of
the complex fraction so that each is a single
fraction.
2) Perform the indicated division by multiplying
the numerator of the complex fraction by the
reciprocal of the denominator of the complex
fraction.
3) Simplify, if possible.
Martin-Gay, Intermediate Algebra, 5ed
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Simplifying Complex Fractions
Example:
x
2
2

x
2
2
x 4
x4

2 2
2  x4 2  x 4
2 x4 x4
x 4
x4

2 2
2
Martin-Gay, Intermediate Algebra, 5ed
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Simplifying Complex Fractions
Simplifying a Complex Fraction (Method 2)
1) Multiply the numerator and the denominator
of the complex fraction by the LCD of the
fractions in both the numerator and
denominator.
2) Simplify, if possible.
Martin-Gay, Intermediate Algebra, 5ed
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Simplifying Complex Fractions
Example:
1 2

2
2
2
y
3 6y
6  4y
 2
2
1 5 6y
6y  5y

y 6
Martin-Gay, Intermediate Algebra, 5ed
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Simplifying with Negative Exponents
When you have a rational expression where
some of the variables have negative
exponents, rewrite the expression using
positive exponents.
The resulting expression is a complex
fraction, so use either method to simplify the
expression.
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Simplifying with Negative Exponents
Example:
x  2 y 2
Simplify 1
1 .
5x  2 y
x 2 y 2

1
1
5x  2 y
1 1
1

2 2
2
2
2 2
x y
1
x y
x y



2
2
5 2 x2 y2
1
1
5
xy

2
x
y

5  2
x y
x
y
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§ 6.4
Dividing Polynomials:
Long Division and
Synthetic Division
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Dividing a Polynomial by a Monomial
Dividing a Polynomial by a Monomial
Divide each term in the polynomial by the
monomial.
ab a b
  , where c  0.
c
c c
Example: Divide
 12a 3  36a  15
 12a 3 36a 15



3a
3a
3a 3a
5
2
 4a  12 
a
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Dividing a Polynomial by a Monomial
4
3
2
10
t

35
t

5
t
Example: Divide
5t 2
10t 4  35t 3  5t 2  10t 4  35t 3  5t 2
5t 2
5t 2
5t 2 5t 2
 2t 2  7t  1
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Dividing Polynomials
Dividing a polynomial by a polynomial other
than a monomial uses a “long division”
technique that is similar to the process known
as long division in dividing two numbers,
which is reviewed on the next slide.
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Dividing Polynomials
168
43 7256
43
295
258
376
344
32
Divide 43 into 72.
Multiply 1 times 43.
Subtract 43 from 72.
Bring down 5.
Divide 43 into 295.
Multiply 6 times 43.
Subtract 258 from 295.
Bring down 6.
Divide 43 into 376.
Multiply 8 times 43.
Subtract 344 from 376.
Nothing to bring down.
We then write our result as
32
168 .
43
Martin-Gay, Intermediate Algebra, 5ed
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Dividing Polynomials
As you can see from the previous example, there is
a pattern in the long division technique.
Divide
Multiply
Subtract
Bring down
Then repeat these steps until you can’t bring
down or divide any longer.
We will incorporate this same repeated technique
with dividing polynomials.
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Dividing Polynomials
4x  5
2
7 x  3 28x  23x  15
2
28x  12x
 35 x  15
 35 x  15
Divide 7x into 28x2.
Multiply 4x times 7x+3.
Subtract 28x2 + 12x from 28x2 – 23x.
Bring down – 15.
Divide 7x into –35x.
Multiply – 5 times 7x+3.
Subtract –35x–15 from –35x–15.
Nothing to bring down.
So our answer is 4x – 5.
Martin-Gay, Intermediate Algebra, 5ed
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Dividing Polynomials
2 x  10
2
2 x  7 4x  6 x  8
2
4 x  14 x
20x  8
20x  70
78
Divide 2x into 4x2.
Multiply 2x times 2x+7.
Subtract 4x2 + 14x from 4x2 – 6x.
Bring down 8.
Divide 2x into –20x.
Multiply -10 times 2x+7.
Subtract –20x–70 from –20x+8.
Nothing to bring down.
We write our final answer as 2 x  10 
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( 2 x  7)
52
Dividing Polynomials Using Long Division
2
c
 3c  2
Example: Divide:
c 1
c
c  1 c 2  3c  2
 c2  c

2. Multiply c by c + 1.

1. Divide the leading term of
the dividend, c2, by the
first term of the divisor, x.
c c2  c
2c
c(c  1)  c 2  c
3. Subtract c2 + c from c2 + 3c – 2.
(c 2  3c)  (c 2  c)  2c
Continued.
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Dividing Polynomials Using Long Division
Example continued:
c 2
c  1 c 2  3c  2
 c2  c

4. Repeat the process until the
degree of the remainder is less
than the degree of the binomial
divisor.

2c 2
  2c  2 
4
Bring down the next term to
obtain a new polynomial.
Remainder
5. Check by verifying that (Quotient)(Divisor) + Remainder = Dividend.
(c  2)(c  1)  (4)
 c 2  3c  2  (4)
 c2  3c  2 
c 2  3c  2  c  2  4
c 1
c 1
Martin-Gay, Intermediate Algebra, 5ed
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Dividing Polynomials Using Long Division
Example: Divide: (y2 – 5y + 6) ÷ (y – 2)
y –3
y  2 y2  5 y  6
y2 – 2y
– 3y + 6
– 3y + 6
0
No remainder
Check : (y – 2)(y – 3) = y2 – 5y + 6 
(y2 – 5y + 6) ÷ (y – 2) = y – 3
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Synthetic Division
To find the quotient and remainder when a polynomial of degree 1
or higher is divided by x – c, a shortened version of long division
called synthetic division may be used.
Long Division
Synthetic Division
x 2
x  1 x 2  3x  2
( x 2  x )
x 2
x  1 x 2  3x  2
x
2x
2
4
2x  2
(2 x  2)
4
Subtraction signs
are not necessary.
The original terms in
red are not needed
because they are the
same as the term
directly above.
Continued.
Martin-Gay, Intermediate Algebra, 5ed
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Synthetic Division
Continued.
x 2
x 1 1 3  2
1
2
2
4
x 2
x  1 x 2  3x  2
x
2x
2
4
x 2
x 1 1 3  2
1 2
2 4
The variables are
not necessary.
The “boxed” numbers
can be aligned
horizontally.
Martin-Gay, Intermediate Algebra, 5ed
Continued.
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Synthetic Division
Continued.
x 2
x 1 1 3  2
1 2
1 2 4
The first two numbers in the last row are the
coefficients of the quotient, the last number is
the remainder.
The leading coefficient of the
dividend can be brought down.
The x + 1
is replaced
with a – 1.
1 1 3  2
1 2
1 2 4
Quotient
To simplify further, the top
row can be removed and
instead of subtracting, we
can change the sign of each
entry and add.
Remainder
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§ 6.5
Solving Equations
Containing Rational
Expressions
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Solving Equations
First note that an equation contains an equal sign
and an expression does not.
To solve EQUATIONS containing rational
expressions, clear the fractions by multiplying
both sides of the equation by the LCD of all the
fractions.
Then solve as in previous sections.
Note: this works for equations only, not
simplifying expressions.
Martin-Gay, Intermediate Algebra, 5ed
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Solving Equations
Example:
Solve the following rational equation.
5
7
1 
3x
6
 5
 7
6 x  1   6 x
 3x   6 
10  6 x  7 x
10  x
Check in the original equation.
5
7
1 
3 10
6
5
7
1 
30
6
1
7
1 
6
6
Martin-Gay, Intermediate Algebra, 5ed
true
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Solving Equations
Example:
Solve the following rational equation.
1
1
1

 2
2 x x  1 3x  3x

1  
1
 1
6 xx  1
6 xx  1 
  
 2 x x  1   3x( x  1) 
3x  1  6 x  2
3x  3  6 x  2
3  3x  2
 3x  1
x 1
3
Martin-Gay, Intermediate Algebra, 5ed
Continued.
62
Solving Equations
Example continued:
Substitute the value for x into the original
equation, to check the solution.
1
1
1


2
1
1
2
1 3 1  3 1
3
3
3
3
   
   
3 3
1
 
2 4 1 1
3
6 3 3
 
4 4 4
true
So the solution is x  13
Martin-Gay, Intermediate Algebra, 5ed
63
Solving Equations
Example:
Solve the following rational equation.
x2
1
1


2
x  7 x  10 3 x  6 x  5
x2
1 

  1
3x  2x  5 2


3x  2x  5
 x  7 x  10   3x  6 x  5 
3x  2  x  5  3x  2
3x  6  x  5  3x  6
3x  x  3x  5  6  6
5 x  7
x  7
5
Martin-Gay, Intermediate Algebra, 5ed
Continued.
64
Solving Equations
Example continued:
Substitute the value for x into the original
equation, to check the solution.

7 2
1
1
5


2
7
7
7
 7  7  10 3  5  6  5  5
5
5
3
1
1
5


49  49  10  21  6 18
5
25
5
5
5 5 5
18 9 18 true
So the solution is x   7 5





Martin-Gay, Intermediate Algebra, 5ed


65
Solving Equations
Example:
Solve the following rational equation.
1
2

x 1 x 1
 1   2 
x  1x  1

x  1x  1
 x 1   x  1 
x  1  2x  1
x 1  2x  2
3 x
Continued.
Martin-Gay, Intermediate Algebra, 5ed
66
Solving Equations
Example continued:
Substitute the value for x into the original
equation, to check the solution.
1
2

3 1 3 1
1 2

2 4
true
So the solution is x = 3.
Martin-Gay, Intermediate Algebra, 5ed
67
Solving Equations
Example:
Solve the following rational equation.
12
3
2


2
9a
3 a 3 a
3   2 
 12
3  a 3  a 


3  a 3  a 
2
 9  a 3 a   3 a 
12  33  a   23  a 
12  9  3a  6  2a
21  3a  6  2a
15  5a
3a
Martin-Gay, Intermediate Algebra, 5ed
Continued.
68
Solving Equations
Example continued:
Substitute the value for x into the original
equation, to check the solution.
12
3
2
2  33  33
93
12 3 2
 
0 5 0
Since substituting the suggested value of a into the
equation produced undefined expressions, the
solution is .
Martin-Gay, Intermediate Algebra, 5ed
69
§ 6.6
Rational Equations
and Problem Solving
Martin-Gay, Intermediate Algebra, 5ed
70
Solving Equations with Multiple Variables
Solving Equations for a Specified Variable
1)
2)
3)
4)
5)
Clear the equation of fractions or rational expressions by
multiplying each side of the equation by the LCD of the
denominators in the equation.
Use the distributive property to remove grouping symbols
such as parentheses.
Combine like terms on each side of the equation.
Use the addition property of equality to rewrite the equation as
an equivalent equation with terms containing the specified
variable on one side and all other terms on the other side.
Use the distributive property and the multiplication property
of equality to get the specified variable alone.
Martin-Gay, Intermediate Algebra, 5ed
71
Solving Equations with Multiple Variables
Example:
Solve the following equation for R1
1 1 1
 
R R1 R2
1 
1  1
RR1 R2       RR1 R2
 R   R1 R2 
R1R2  RR2  RR1
R1R2  RR1  RR2
R1 R2  R  RR2
RR2
R1 
R2  R
Martin-Gay, Intermediate Algebra, 5ed
72
Finding an Unknown Number
Example:
The quotient of a number and 9 times its reciprocal
is 1. Find the number.
1.) Understand
Read and reread the problem. If we let
n = the number, then
1
= the reciprocal of the number
n
Continued
Martin-Gay, Intermediate Algebra, 5ed
73
Finding an Unknown Number
Example continued:
2.) Translate
The quotient of
is
a number
and 9 times its reciprocal
n
1
9 
n

1
=
1
Continued
Martin-Gay, Intermediate Algebra, 5ed
74
Finding an Unknown Number
Example continued:
3.) Solve
 1
n  9   1
 n
9
n   1
n
n
n 1
9
n2  9
n  3,3
Martin-Gay, Intermediate Algebra, 5ed
Continued
75
Finding an Unknown Number
Example continued:
4.) Interpret
Check: We substitute the values we found from the
equation back into the problem. Note that nothing in
the problem indicates that we are restricted to positive
values.
 1
 1 
3  9    1
 3  9
 1
 3
 3
3  3  1 true
 3  3  1 true
State: The missing number is 3 or –3.
Martin-Gay, Intermediate Algebra, 5ed
76
Ratios and Rates
Ratio is the quotient of two numbers or two
quantities.
The ratio of the numbers a and b can also be
a
written as a:b, or .
b
The units associated with the ratio are important.
The units should match.
If the units do not match, it is called a rate, rather
than a ratio.
Martin-Gay, Intermediate Algebra, 5ed
77
Proportions
Proportion is two ratios (or rates) that are
equal to each other.
a c

b d
We can rewrite the proportion by multiplying
by the LCD, bd.
This simplifies the proportion to ad = bc.
This is commonly referred to as the cross product.
Martin-Gay, Intermediate Algebra, 5ed
78
Solving Proportions
Example:
Solve the proportion for x.
x 1 5

x2 3
3x  1  5x  2
3x  3  5 x  10
 2x  7
x  7
2
Martin-Gay, Intermediate Algebra, 5ed
Continued.
79
Solving Proportions
Example continued:
Substitute the value for x into the original
equation, to check the solution.
 7 1 5
2

7 2 3
2
5
25
3
3
2
true
So the solution is x   7 2
Martin-Gay, Intermediate Algebra, 5ed
80
Solving Proportions
Example:
If a 170-pound person weighs approximately 65 pounds
on Mars, how much does a 9000-pound satellite weigh?
170- pound personon Earth 65 - pound personon Mars

9000- poundsatelliteon Earth x - poundsatelliteon Mars
170x  9000 65  585,000
x  585000/ 170  3441pounds
Martin-Gay, Intermediate Algebra, 5ed
81
Solving Proportions
Example:
Given the following prices charged for
various sizes of picante sauce, find the best
buy.
• 10 ounces for $0.99
• 16 ounces for $1.69
• 30 ounces for $3.29
Continued.
Martin-Gay, Intermediate Algebra, 5ed
82
Solving Proportions
Example continued:
Size
Price
Unit Price
10 ounces
$0.99
$0.99/10 = $0.099
16 ounces
$1.69
$1.69/16 = $0.105625
30 ounces
$3.29
$3.29/30  $0.10967
The 10 ounce size has the lower unit price, so it is the
best buy.
Martin-Gay, Intermediate Algebra, 5ed
83
Solving a Work Problem
Example:
An experienced roofer can roof a house in 26 hours. A
beginner needs 39 hours to do the same job. How long will it
take if the two roofers work together?
1.) Understand
Read and reread the problem. By using the times for each
roofer to complete the job alone, we can figure out their
corresponding work rates in portion of the job done per hour.
Time in hrs
Experienced roofer 26
Beginner roofer
39
Together
t
Portion job/hr
1/26
/39
1/t
Martin-Gay, Intermediate Algebra, 5ed
Continued
84
Solving a Work Problem
Example continued:
2.) Translate
Since the rate of the two roofers working together
would be equal to the sum of the rates of the two
roofers working independently,
1
1 1


26 39 t
Continued
Martin-Gay, Intermediate Algebra, 5ed
85
Solving a Work Problem
Example continued:
3.) Solve
1
1 1


26 39 t
1  1
 1
78t      78t
 26 39   t 
3t  2t  78
5t  78
t  78 / 5 or 15.6 hours
Continued
Martin-Gay, Intermediate Algebra, 5ed
86
Solving a Work Problem
Example continued:
4.) Interpret
Check: We substitute the value we found from the
proportion calculation back into the problem.
1
1
1


26 39 78
3
2
5


78 78 78
5
true
State: The roofers would take 15.6 hours working
together to finish the job.
Martin-Gay, Intermediate Algebra, 5ed
87
Solving a Rate Problem
Example:
The speed of Lazy River’s current is 5 mph. A boat travels 20
miles downstream in the same time as traveling 10 miles
upstream. Find the speed of the boat in still water.
1.) Understand
Read and reread the problem. By using the formula d=rt, we
can rewrite the formula to find that t = d/r.
We note that the rate of the boat downstream would be the rate
in still water + the water current and the rate of the boat
upstream would be the rate in still water – the water current.
Distance rate time = d/r
Down 20
r + 5 20/(r + 5)
Up
10
r – 5 10/(r – 5)
Continued
Martin-Gay, Intermediate Algebra, 5ed
88
Solving a Rate Problem
Example continued:
2.) Translate
Since the problem states that the time to travel
downstairs was the same as the time to travel
upstairs, we get the equation
20
10

r 5 r 5
Continued
Martin-Gay, Intermediate Algebra, 5ed
89
Solving a Rate Problem
Example continued:
3.) Solve
20
10

r 5 r 5
 20   10 
r  5r  5

r  5r  5
 r 5  r 5
20r  5  10r  5
20 r  100  10 r  50
10 r  150
r  15 mph
Martin-Gay, Intermediate Algebra, 5ed
Continued
90
Solving a Rate Problem
Example continued:
4.) Interpret
Check: We substitute the value we found from the
proportion calculation back into the problem.
20
10

15  5 15  5
20 10

true
20 10
State: The speed of the boat in still water is 15 mph.
Martin-Gay, Intermediate Algebra, 5ed
91
§ 6.7
Variation and
Problem Solving
Martin-Gay, Intermediate Algebra, 5ed
92
Direct Variation
y varies directly as x, or y is directly proportional to
x, if there is a nonzero constant k such that y = kx.
The family of equations of the form y = kx are
referred to as direct variation equations.
The number k is called the constant of variation or
the constant of proportionality.
Martin-Gay, Intermediate Algebra, 5ed
93
Direct Variation
If y varies directly as x, find the constant of
variation k and the direct variation equation,
given that y = 5 when x = 30.
y = kx
5 = k·30
k = 1/6
1
So the direct variation equation is y =
x
6
Martin-Gay, Intermediate Algebra, 5ed
94
Direct Variation
Example:
If y varies directly as x, and y = 48 when x = 6, then
find y when x = 15.
y = kx
48 = k·6
8=k
So the equation is y = 8x.
y = 8·15
y = 120
Martin-Gay, Intermediate Algebra, 5ed
95
Direct Variation
Example:
At sea, the distance to the horizon is directly
proportional to the square root of the elevation
of the observer. If a person who is 36 feet
above water can see 7.4 miles, find how far a
person 64 feet above the water can see.
Round your answer to two decimal places.
Continued.
Martin-Gay, Intermediate Algebra, 5ed
96
Direct Variation
Example continued:
We substitute our given value
for the elevation into the
equation.
d k e
7.4  k 36
7.4  6 k
7.4
k
6
So our equation is
7 .4
d
64
6
7.4
59.2
d
(8) 
 9.87 miles
6
6
7.4
d
e
6
Martin-Gay, Intermediate Algebra, 5ed
97
Inverse Variation
y varies inversely as x, or y is inversely proportional
to x, if there is a nonzero constant k such that y = k/x.
The family of equations of the form y = k/x are
referred to as inverse variation equations.
The number k is still called the constant of variation
or the constant of proportionality.
Martin-Gay, Intermediate Algebra, 5ed
98
Inverse Variation
Example:
If y varies inversely as x, find the constant of
variation k and the inverse variation equation,
given that y = 63 when x = 3.
y = k/x
63 = k/3
k = 63·3
k = 189
189
So the inverse variation equation is y =
x
Martin-Gay, Intermediate Algebra, 5ed
99
Powers of x
y can vary directly or inversely as powers of x,
as well.
y varies directly as a power of x if there is a
nonzero constant k and a natural number n
such that y = kxn
y varies inversely as a power of x if there is a
nonzero constant k and a natural number n
such that y  k
x
n
Martin-Gay, Intermediate Algebra, 5ed
100
Powers of x
Example:
The maximum weight that a circular column
can hold is inversely proportional to the
square of its height.
If an 8-foot column can hold 2 tons, find how
much weight a 10-foot column can hold.
Continued.
Martin-Gay, Intermediate Algebra, 5ed
101
Powers of x
Example continued:
k
w 2
h
k
k
2 2 
8
64
We substitute our given value
for the height of the column
into the equation.
128 128
w 2 
 1.28 tons
10
100
k  128
So our equation is
128
w 2
h
Martin-Gay, Intermediate Algebra, 5ed
102
Combined Variation
y varies jointly as, or is jointly proportional to
two (or more) variables x and z, if there is a
nonzero constant k such that y = kxz.
k is STILL the constant of variation or the
constant of proportionality.
We can also use combinations of direct,
inverse, and joint variation. These variations
are referred to as combined variations.
Martin-Gay, Intermediate Algebra, 5ed
103
Combined Variation
Example:
Write the following statements as equations.
• a varies jointly as b and c
a = kbc
• P varies jointly as R and the square of S
P = kRS2
• The weight (w) varies jointly with the width (d)
and the square of the height (h) and inversely
with the length (l)
kdh2
w
l
Martin-Gay, Intermediate Algebra, 5ed
104