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Bruce Mayer, PE Chabot College Mathematics
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
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Chabot Mathematics
§6.6 Rational§6.6 RationalEquationsEquations
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Bruce Mayer, PE Chabot College Mathematics
Review §Review §
Any QUESTIONS About• §6.4 → Complex Rational Expressions
Any QUESTIONS About HomeWork• §6.4 → HW-21
6.4 MTH 55
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Bruce Mayer, PE Chabot College Mathematics
Solving Rational EquationsSolving 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:
2
2 7 5 8 6 2 58, , 4.
4 3 3 9
x xx
x x x x x x
We want determine the value(s) for x that make these Equations TRUE
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Bruce Mayer, PE Chabot College Mathematics
To Solve a Rational EquationTo Solve a Rational Equation1. 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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Bruce Mayer, PE Chabot College Mathematics
Example Example Solve Solve
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
1
5 2 4
x x
20 201
5 2 4
x x
20 204
01
22
5
x x
4 10 5x x 6 5x
5
6x
Using the multiplication principle to multiply both sides by the LCM. Parentheses are important!
Using the distributive law. Be sure to multiply EACH term by the LCM
Simplifying and solving for x. If fractions remain, we have either made a mistake or have not used the LCM of ALL the denominators.
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Bruce Mayer, PE Chabot College Mathematics
Checking AnswersChecking 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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Bruce Mayer, PE Chabot College Mathematics
Example Example Solve Solve
SOLUTION - Note that x canNOT equal 0. The Denominator LCM is 15x.
1 1 4
3 15x x
1 1 4
3 1515 15
x xx x
5 15 4x
20 4x
5 x
515x1
3x 15 x
1
x 15
4
15x
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Bruce Mayer, PE Chabot College Mathematics
Example Example Solve Solve
CHECKtentative Solution,x = 5
1 1 4
3 15x x
1 1 4
3 15x x
1 1 4
3( ) 15
1 1 4
15 5 151 3 4
15 15 154 4
15 15
5 5
The Solutionx = 5 CHECKS
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Bruce Mayer, PE Chabot College Mathematics
Example Example Solve Solve
SOLUTION - Note that x canNOT equal 0. The Denom LCM is x
127x
x
12(7)x
xx x
x xx 12
x 7x
2 12 7x x 2 7 12 0x x
( 3)( 4) 0x x ( 3) 0 or ( 4) 0x x
Thus by Zero Products:
x = 3 or
x = 4
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Bruce Mayer, PE Chabot College Mathematics
Example Example Solve Solve
CHK: For x = 3 For x = 4
127x
x
127x
x
127x
x
312
73
3 4 7
7 7
412
74
4 3 7
7 7
Both of these check, so there are two solutions; 3 and 4
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Bruce Mayer, PE Chabot College Mathematics
Example Example Solve Solve
SOLUTION Note that y canNOT equal 3 or −3. We multiply both sides of the equation by the Denom LCM.
5 3 2
( 3)( 3( 3)( 3) ( 3 ( 3
) 3 3) )y
y y y yy y y
( 3y )( 3y )5
( 3y )( 3y )
( 3y
)( 3)3
3
y
y
( 3)( 3y y
)2
3y
5 3( 3) 2( 3)y y
3 9 25 6y y
5 15y
20 y
2
5 3 2
9 3 3y y y
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Bruce Mayer, PE Chabot College Mathematics
Example Example Solve Solve
SOLUTION - Note that x canNOT equal 1 or −1. Multiply both sides of the eqn by the LCM
( 1)( 1) ( 1)( 12 5 4
1 1 ( 1)( 1)
)x x x x
x x x x
2( 1) 5( 1) 4x x
2 2 5 5 4x x
3 7 4x
3 3x
1x
Because of the restriction above, 1 must be rejected as a solution. This equation has NO solution.
11
4
1
5
1
2
xxxx
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Bruce Mayer, PE Chabot College Mathematics
Example Example Solve Solve
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:
2 7 58.
4
x x
x x
Multiplying by the LCD to clear fractions
2 784 4
5
4
x x
x xx x
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Bruce Mayer, PE Chabot College Mathematics
Example Example Solve Solve
SOLN cont.
2 7 58.
4
x x
x x
2 7 544 84
4x x
x xx
x x
(2 7) 4 ( 5)32
4
4
x xx
x x
x x
Using the distributive law
Locating factors equal to 1
Removing factors equal to 1
Using the distributive law
(2 7) 4( 5) 32x x x
2 7 4 20 32x x x
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Bruce Mayer, PE Chabot College Mathematics
Example Example Solve Solve
SOLN cont.
2 7 58.
4
x x
x x
This should check since x 0.
6 13 32x x
13 26x12 x
CHECK 2 7 58
4
x x
x x
8 8
1 12 2
1 12 2
2 7 5
4
3 11
8
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Bruce Mayer, PE Chabot College Mathematics
Rational Eqn Rational Eqn CAUTIONCAUTION
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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Bruce Mayer, PE Chabot College Mathematics
Example Example Solve Solve
SOLUTION: To find all restrictions and to assist in finding the LCD, we factor:
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
2
8 6 2.
3 3 9x x x
8 6 2.
3 3 ( 3)( 3)x x x x
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Bruce Mayer, PE Chabot College Mathematics
Example Example Solve Solve
SOLUTION: By LCD Multiplication
Remove factors Equal to One and solve the resulting Eqn• Keep in Mind any restrictions
2
8 6 2.
3 3 9x x x
8 6 2
3 3 ( 3)( 3)( 3) ( 3)( 3)
( 3)x x x xx x x x
( 3)( 3) ( 3)8 6 ( 3)( 3)2
( 3)3 3 ( 3)( 3)
x x x xx x
x x x x
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Bruce Mayer, PE Chabot College Mathematics
Example Example Solve Solve
SOLN cont.: Multiply and Collect Similar terms
A check will confirm that 22 is the solution
2
8 6 2.
3 3 9x x x
8( 3) 6( 3) 2x x
8 24 6 18 2x x
2 42 2x
2 44x
22x
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Bruce Mayer, PE Chabot College Mathematics
Example Example Eqn with NO Soln Eqn with NO Soln
To avoid division by zero, exclude from the expression domain
1 and –1, since these values make one or more of the denominators in the
equation equal 0.
Distributive property
Solve .3x – 1
=2x + 1
– 6
x2 – 1
=3
x – 12
x + 1– 6
x2 – 1(x – 1)(x + 1)(x – 1)(x + 1)
=3
x – 12
x + 1– 6
x2 – 1(x – 1)(x + 1)(x – 1)(x + 1) (x – 1)(x + 1)
=– 63(x + 1) 2(x – 1)
=– 63x + 3 2x + 2
= 6x + 5
= 1x
Multiply each side by the LCD, (x –1)(x + 1).
Multiply.
Distributive property
Combine terms.
Subtract 5.
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Bruce Mayer, PE Chabot College Mathematics
Example Example Eqn with NO Soln Eqn with NO SolnSolve .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 – 12
x + 1– 6
x2 – 1
=3
1 – 12
1 + 1– 6
12 – 1
=30
22
– 60
Since division by 0 is undefined, the given equation has no solution,
and the solution set is ∅.
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Bruce Mayer, PE Chabot College Mathematics
Example Example Fcn to Eqn Fcn to Eqn
Given Function:
Find all values of a for which
5( ) .f x x
x
( ) 4.f a
SOLUTION On BoardOn Board By Function Notation:
5( )f a a
a
Thus Need to find all values of a for which f(a) = 4
54.a
a
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Bruce Mayer, PE Chabot College Mathematics
Example Example Fcn to Eqn Fcn to Eqn
Solve for a:
First note that a 0. To solve for a, multiply both sides of the equation by the LCD, a:
54.a
a
Multiplying both sides by a. Parentheses are important.
54 aa
aa
54a a
aa a Using the distributive law
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Bruce Mayer, PE Chabot College Mathematics
Example Example Fcn to Eqn Fcn to Eqn
CarryOutSolution
CHECK
Simplifying2 5 4a a
2 4 5 0a a
( 5)( 1) 0a a
5 1a or a
Getting 0 on one side
Factoring
Using the principle of zero products
5( ) 5 1 4;
5( ) 1
5 55
1 1 41
5 .
f
f
STATE: The solutions are 5 and −1. For a = 5 or a = −1, we have f(a) = 4.
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Bruce Mayer, PE Chabot College Mathematics
Rational Equations and GraphsRational 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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Bruce Mayer, PE Chabot College Mathematics
Rational Equations and GraphsRational Equations and Graphs
Graph the function, and on the same grid graph y = g(x) = 4
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.
4
-1 5
5( )f x x
x
4y
x
y
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Bruce Mayer, PE Chabot College Mathematics
Rational Equations and GraphsRational 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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Bruce Mayer, PE Chabot College Mathematics
WhiteBoard WorkWhiteBoard Work
Problems From §6.6 Exercise Set• 34, 38, 62
Rational Expressions
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Bruce Mayer, PE Chabot College Mathematics
All Done for TodayAll Done for Today
Remember:can NOTDivide by
ZERO
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Bruce Mayer, PE Chabot College Mathematics
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
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Chabot Mathematics
AppendiAppendixx
–
srsrsr 22
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Bruce Mayer, PE Chabot College Mathematics
Graph Graph yy = | = |xx||
Make T-tablex y = |x |
-6 6-5 5-4 4-3 3-2 2-1 10 01 12 23 34 45 56 6
x
y
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
file =XY_Plot_0211.xls
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Bruce Mayer, PE Chabot College Mathematics
x
y
-3
-2
-1
0
1
2
3
4
5
-3 -2 -1 0 1 2 3 4 5
M55_§JBerland_Graphs_0806.xls -10
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
1
2
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
file =XY_Plot_0211.xls
xy