Chabot Mathematics
§6.6 Rational
Equations
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-34_sec_6-6_Rational_Equations.ppt
Review § 6.4
MTH 55
 Any QUESTIONS About
• §6.4 → Complex Rational Expressions
 Any QUESTIONS About HomeWork
• §6.4 → HW-26
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-34_sec_6-6_Rational_Equations.ppt
Solving Rational Equations
 In previous sections, we learned how to
simplify expressions. We now learn to
solve a new type of equation. A
rational equation is an equation that
contains one or more rational
expressions. Some examples:
2x  7 x  5
8
6
2
5

 8,

 2
, x   4.
4x
x
x 3 x 3 x 9
x
 We want determine the value(s) for x
that make these Equations TRUE
Chabot College Mathematics
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-34_sec_6-6_Rational_Equations.ppt
To Solve a Rational Equation
1. List any restrictions that exist.
Numbers that make a denominator equal 0
canNOT possibly be solutions.
2. CLEAR the equation of FRACTIONS by
multiplying both sides by the LCM of ALL
the denominators present
3. Solve the resulting equation using the
addition principle, the multiplication principle,
and the Principle of Zero Products, as
needed.
4. Check the possible solution(s) in the
original equation.
Chabot College Mathematics
4
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-34_sec_6-6_Rational_Equations.ppt
Example  Solve
x x 1
 
5 2 4
 SOLUTION - Because no variable appears in the
denominator, no restrictions exist. The LCM of 5,
2, and 4 is 20, so we multiply both sides by 20
Using the multiplication principle
 x
 x 1
20     20    
to multiply both sides by the LCM.
Parentheses are important!
5
2 4
x
x
1
Using the distributive law.
20   20   20 
Be sure to multiply
5
2
4
EACH term by the LCM
4x  10x  5
Simplifying and solving for x. If
6x  5
fractions remain, we have either
made a mistake or have not used
5
the LCM of ALL the denominators.
x
6
Chabot College Mathematics
5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-34_sec_6-6_Rational_Equations.ppt
Checking Answers
 Since a variable expression could
represent 0, multiplying both sides
of an equation by a variable
expression does NOT always
produce an Equivalent Equation
• COULD be Multiplying by Zero and
Not Know it
 Thus checking each solution in
the original equation is essential.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-34_sec_6-6_Rational_Equations.ppt
1 1 4
 
Example  Solve
3 x x 15
 SOLUTION - Note that x canNOT
equal 0. The Denominator LCM is 15x.
4
 1 1
15 x     15 x 
15
 3x x 
1
1
4
5
15x 
 15 x   15 x 
3x
x
15
5 15  4x
20  4x
5 x
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-34_sec_6-6_Rational_Equations.ppt
Example  Solve
 CHECK
tentative
Solution,
x=5
 The Solution
x = 5 CHECKS
Chabot College Mathematics
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1 1 4
 
3 x x 15
1 1 4
 
3 x x 15
1
1 4
 
3(5) 5 15
1 1 4
 
15 5 15
1 3
4
 
15 15 15
4
4

15 15

Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-34_sec_6-6_Rational_Equations.ppt
Example  Solve
12
x 7
x
 SOLUTION - Note that x canNOT
equal 0. The Denom LCM is x
12 

x  x    x(7)
x

12
x  x  x   7x
x
x 2  12  7 x
x 2  7 x  12  0
 Thus by Zero
Products:
x=3
or
x=4
( x  3)( x  4)  0
( x  3)  0 or
( x  4)  0
Chabot College Mathematics
9
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-34_sec_6-6_Rational_Equations.ppt
Example  Solve
 CHK: For x = 3
12
x 7
x
For x = 4
12
x 7
x
12
3  7
3
12
x 7
x
12
4  7
4
3 4  7
43  7
77
77
 Both of these check, so there are
two solutions; 3 and 4
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-34_sec_6-6_Rational_Equations.ppt
Example  Solve
5
3
2


2
y 9 y 3 y 3
 SOLUTION  Note that y canNOT
equal 3 or −3. We multiply both sides of
the equation by the Denom LCM.


 3
5
2 
( y  3)( y  3) 

  ( y  3)( y  3) 

 ( y  3)( y  3) 
 y 3 y 3
( y  3 )( y  3 )5
( y  3 )( y  3)3 ( y  3)( y  3 )2


( y  3 )( y  3 )
y 3
y 3
5  3( y  3)  2( y  3)
5  y  15
5  3y  9  2 y  6
20  y
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-34_sec_6-6_Rational_Equations.ppt
Example  Solve
2
5
4


x  1 x  1 x  1x  1
 SOLUTION - Note that x canNOT equal 1 or
−1. Multiply both sides of the eqn by the LCM
5 
4
 2
( x  1)( x  1) 

( x  1)( x  1)

 x  1 x  1  ( x  1)( x  1)
2( x  1)  5( x  1)  4
2 x  2  5x  5  4
3x  7  4
3x  3
x 1
Chabot College Mathematics
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Because of the restriction
above, 1 must be rejected
as a solution. This equation
has NO solution.
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-34_sec_6-6_Rational_Equations.ppt
Example  Solve
2x  7 x  5

 8.
4x
x
 SOLUTION: Because the left side of
this equation is undefined when x is 0,
we state at the outset that x  0.
 Next, we multiply both sides of the
equation by the LCD, 4x:
 2x  7 x  5 
4x 

  4x  8
x 
 4x
Chabot College Mathematics
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Multiplying by the LCD
to clear fractions
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-34_sec_6-6_Rational_Equations.ppt
Example  Solve
2x  7 x  5

 8.
4x
x
 SOLN cont.
2x  7
x5
4x 
 4x 
 4x 8
4x
x
Using the
distributive law
4 x(2 x  7) 4  x( x  5)

 32 x
4x
x
Locating factors
equal to 1
(2 x  7)  4( x  5)  32 x
Removing
factors equal to 1
2x  7  4x  20  32x
Chabot College Mathematics
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Using the
distributive law
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-34_sec_6-6_Rational_Equations.ppt
Example  Solve
 SOLN
cont.
6x 13  32x
13  26x
1
2
 CHECK
2x  7 x  5

 8.
4x
x
x
This should check since x  0.
2x  7 x  5

8
4x
x
2  12  7 12  5
 1
1
4 2
2
8
3  11
Chabot College Mathematics
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88
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-34_sec_6-6_Rational_Equations.ppt
Rational Eqn CAUTION
 When solving rational equations, be
sure to list any restrictions as part
of the first step.
 Refer to the restriction(s) as you
proceed
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-34_sec_6-6_Rational_Equations.ppt
Example  Solve
8
6
2

 2
.
x 3 x 3 x 9
 SOLUTION: To find all restrictions and
to assist in finding the LCD, we factor:
8
6
2


.
x  3 x  3 ( x  3)( x  3)
 Note that to prevent division by zero
x  3 and x  −3.
 Next multiply by the LCD, (x + 3)(x – 3),
and then use the distributive law
Chabot College Mathematics
17
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-34_sec_6-6_Rational_Equations.ppt
Example  Solve
8
6
2

 2
.
x 3 x 3 x 9
 SOLUTION: By LCD Multiplication
6 
2
 8
( x  3)( x  3) 

  ( x  3)( x  3) 
( x  3)( x  3)
 x 3 x 3
8
6
( x  3)( x  3)2
( x  3)( x  3) 
 ( x  3)( x  3) 

x3
x  3 ( x  3)( x  3)
 Remove factors Equal to One and solve
the resulting Eqn
• Keep in Mind any restrictions
Chabot College Mathematics
18
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-34_sec_6-6_Rational_Equations.ppt
Example  Solve
 SOLN cont.:
Multiply and
Collect
Similar terms
8
6
2

 2
.
x 3 x 3 x 9
8( x  3)  6( x  3)  2
8x  24  6x 18  2
2x  42  2
2x  44
x  22
 A check will confirm that 22 is the solution
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-34_sec_6-6_Rational_Equations.ppt
Example  Eqn with NO Soln
Solve
3
x–1
–
2
x+1
=
6 .
x2 – 1
Multiply
each
side by
LCD,
(x –1)(x
+ 1).
To avoid
division
bythe
zero,
exclude
from
the expression domain
1 and –1, since these values make
one or2more of the denominators 6
in the
3
–
(x – 1)(x + 1)
= (x – 1)(x + 1) 2
equation equal 0.
x–1
x+1
x –1
(x – 1)(x + 1)
3
x–1
Chabot College Mathematics
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– (x – 1)(x + 1)
3(x + 1)
–
3x + 3
–
2
x+1
=
(x – 1)(x + 1)
6
x2 – 1
2(x
– 1) = property
6 Multiply.
Distributive
2x + 2
=
6
Distributive property
x+5
=
6
Combine terms.
x
=
1
Subtract 5.
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-34_sec_6-6_Rational_Equations.ppt
Example  Eqn with NO Soln
Solve
3
x–1
–
2
x+1
=
6 .
x2 – 1
Since 1 is not in the domain, it cannot be a solution of the equation.
Substituting 1 in the original equation shows why.
Check:
3
x–1
–
2
x+1
=
6
x2 – 1
3
1–1
–
2
1+1
=
6
12 – 1
3
0
–
2
2
=
6
0
Since division by 0 is undefined, the given equation has no solution,
and the solution set is ∅.
Chabot College Mathematics
21
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-34_sec_6-6_Rational_Equations.ppt
Example  Fcn to Eqn
5
 Given Function: f ( x)  x  .
x
 Find all values of a for which f (a)  4.
 SOLUTION  On Board
5
 By Function Notation: f (a)  a  a
 Thus Need to find all values of a for
which f(a) = 4
5
a   4.
a
Chabot College Mathematics
22
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-34_sec_6-6_Rational_Equations.ppt
Example  Fcn to Eqn
5
 Solve for a: a   4.
a
 First note that a  0. To solve for a,
multiply both sides of the equation by
the LCD, a:
5

aa    4a
a

Multiplying both sides by a.
Parentheses are important.
5
a  a  a   4a Using the distributive law
a
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-34_sec_6-6_Rational_Equations.ppt
Example  Fcn to Eqn
 CarryOut
Solution
 CHECK
a 2  5  4a
a 2  4a  5  0
Simplifying
Getting 0 on one side
(a  5)(a  1)  0
Factoring
a  5 or
Using the principle of
zero products
a  1
5
 5  1  4;
5
5
f (1)  1 
  1  5  4.
1
f (5)  5 
 STATE: The solutions are 5 and −1. For
a = 5 or a = −1, we have f(a) = 4.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-34_sec_6-6_Rational_Equations.ppt
Rational Equations and Graphs
 One way to visualize the solution to the
last example is to make a graph. This
can be done by graphing; e.g., Given
5
f ( x)  x  .
x
 Find x such that f(x) = 4
Chabot College Mathematics
25
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-34_sec_6-6_Rational_Equations.ppt
Rational Equations and Graphs
 Graph the function,
and on the same
grid graph
y = g(x) = 4
y
y4
4
x
-1
5
5
f ( x)  x 
x
 We then inspect
the graph for any
x-values that are paired with 4. It
appears from the graph that f(x) = 4
when x = 5 or x = −1.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-34_sec_6-6_Rational_Equations.ppt
Rational Equations and Graphs
 Graphing gives approximate solutions
 Although making a graph is not the
fastest or most precise method of
solving a rational equation, it provides
visualization and is useful when
problems are too difficult to solve
algebraically
Chabot College Mathematics
27
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-34_sec_6-6_Rational_Equations.ppt
WhiteBoard Work
 Problems From §6.6 Exercise Set
• 34, 38, 62

Rational
Expressions
Chabot College Mathematics
28
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-34_sec_6-6_Rational_Equations.ppt
All Done for Today
Remember:
can NOT
Divide by
ZERO
Chabot College Mathematics
29
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-34_sec_6-6_Rational_Equations.ppt
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
–
Chabot College Mathematics
30
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-34_sec_6-6_Rational_Equations.ppt
Graph y = |x|
6
 Make T-table
x
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
Chabot College Mathematics
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5
y = |x |
6
5
4
3
2
1
0
1
2
3
4
5
6
y
4
3
2
1
x
0
-6
-5
-4
-3
-2
-1
0
1
2
3
-1
-2
-3
-4
-5
file =XY_Plot_0211.xls
-6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-34_sec_6-6_Rational_Equations.ppt
4
5
6
y
5
2
y
1
x
4
0
-6
-5
-4
-3
-2
-1
3
0
1
2
3
4
-1
-2
2
-3
1
-4
x
-5
5
-6
0
-3
-2
-1
0
1
2
3
-1
4
-7
-8
-2
-9
M55_§JBerland_Graphs_0806.xls
-3
Chabot College Mathematics
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file =XY_Plot_0211.xls
-10
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-34_sec_6-6_Rational_Equations.ppt
5
6