Transcript MTH55A_Lec-24_sec_6-6_Rational_Equations

Chabot Mathematics
§6.6 Rational
Equations
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
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Review § 6.4
MTH 55
 Any QUESTIONS About
• §6.4 → Complex Rational Expressions
 Any QUESTIONS About HomeWork
• §6.4 → HW-21
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Solving Rational Equations
 In previous Lectures, 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
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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.
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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
6 x  5
fractions remain, we have either
made a mistake or have not used
5
the LCM of ALL the denominators.
x
6
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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.
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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
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Example  Solve
 CHECK
tentative
Solution,
x=5
 The Solution
x = 5 CHECKS
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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

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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
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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
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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
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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
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Because of the restriction
above, 1 must be rejected
as a solution. This equation
has NO solution.
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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
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Multiplying by the LCD
to clear fractions
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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
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Using the
distributive law
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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
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88
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Rational Eqn CAUTION
 When solving rational equations, be
sure to list any Division-by-Zero
restrictions as part of the first step.
 Refer to the
restriction(s) as
you proceed
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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
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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
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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
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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
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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.
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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 ∅.
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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
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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
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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.
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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
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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.
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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
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WhiteBoard Work
 Problems From §6.6 Exercise Set
• 34, 38, 62

Rational
Expressions
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All Done for Today
Remember:
can NOT
Divide by
ZERO
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Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
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Licensed Electrical & Mechanical Engineer
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–
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Graph y = |x|
6
 Make T-table
x
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
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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
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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
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file =XY_Plot_0211.xls
-10
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5
6