Chapter Six 6.6

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§ 6.6
Rational Equations
Solving a Rational Equation
A rational equation, also called a fractional equation, is an equation
containing one or more rational expressions. The following is an
example of a rational equation:
7x  4 9 4
  .
5x
5 x
Do you see that there is a variable in the denominator? This is a
characteristic of many rational equations.
P 441
Blitzer, Intermediate Algebra, 5e – Slide #2 Section 6.6
Solving a Rational Equation
When adding or subtracting rational expressions,
we find the LCD and convert fractions to
equivalent fractions that have the common
denominator.
See Study Tip on page 443 for
a comparison of the two
procedures.
By contrast, when we solve rational equations, the
LCD is used as a multiplier that clears an equation
of fractions.
Thus in an equation – you can say…. “I really don’t like these fractions” and then
just proceed to clear them out!
Blitzer, Intermediate Algebra, 5e – Slide #3 Section 6.6
P 442
Solving a Rational Equation
EXAMPLE
Solve: 7 x  4  9  4 .
5x
5
x
SOLUTION
Notice that the variable x appears in two of the denominators. We
must avoid any values of the variable that make a denominator
zero.
7x  4 9 4
 
5x
5 x
This denominator would
equal zero if x = 0.
This denominator would
equal zero if x = 0.
Therefore, we see that x cannot equal zero.
Blitzer, Intermediate Algebra, 5e – Slide #4 Section 6.6
P 442
Solving a Rational Equation
CONTINUED
The denominators are 5x, 5, and x. The least common
denominator is 5x. We begin by multiplying both sides of the
equation by 5x. We will also write the restriction that x cannot
equal zero to the right of the equation.
7x  4 9 4
  , x0
5x
5 x
5x 
7x  4
9 4
 5 x  
5x
5 x
5x 7 x  4 5x 9 5x 4


  
1
5x
1 5 1 x
Blitzer, Intermediate Algebra, 5e – Slide #5 Section 6.6
This is the given equation.
Multiply both sides by 5x,
the LCD.
Use the distributive property.
Solving a Rational Equation
CONTINUED
7x  4  9x  5  4
Divide out common factors
in the multiplications.
7 x  4  9 x  20
Multiply.
 4  2 x  20
Subtract.
16  2 x
Add.
8 x
Divide.
The proposed solution, 8, is not part of the restriction x  0 . It
should check in the original equation.
Blitzer, Intermediate Algebra, 5e – Slide #6 Section 6.6
Solving a Rational Equation
CONTINUED
Check 8:
7x  4 9 4
 
5x
5 x
7 8  4 ? 9 4
 
58
5 8
56  4 ? 9 4
 
40
5 8
52 ? 9 4
 
40 5 8
52 ?
9 4
40 
 40    
40
5 8
9
4
52 ? 40   40 
5
8
52 ? 8  9  5  4
52 ? 72  20
52  52 true
This true statement verifies
that the solution is 8 and the
solution set is {8}.
Blitzer, Intermediate Algebra, 5e – Slide #7 Section 6.6
Solving a Rational Equation
Check Point 1
Solve: x  6  x  24  2
2x
5x
SOLUTION
The denominators are 2x, 5x, and 1. The least common
denominator is 10x and x cannot equal zero.
x  6 x  24

 2, x  0
2x
5x
5
2
10 x x  6 10 x x  24



 10 x  2
1
2x
1
5x
Divide out common factors in the
multiplications.
5 x  30  2 x  48  20 x
Multiply.
Blitzer, Intermediate Algebra, 5e – Slide #8 Section 6.6
P 442
Solving a Rational Equation
CONTINUED
5 x  30  2 x  48  20 x
7 x  78  20 x
78  13 x
Combine like terms.
Subtract.
Divide by 13.
6x
The proposed solution, 6, is not part of the restriction
the original equation.
Blitzer, Intermediate Algebra, 5e – Slide #9 Section 6.6
x  0 . It should check in
Solving a Rational Equation
Solving Rational Equations
1) List restrictions on the variable. Avoid any values of
the variable that make a denominator zero.
2) Clear the equation of fractions by multiplying both
sides by the LCD of all rational expressions in the
equation.
3) Solve the resulting equation.
4) Reject any proposed solution that is in the list of
restrictions on the variable. Check other proposed
solutions in the original equation.
P 442
Blitzer, Intermediate Algebra, 5e – Slide #10 Section 6.6
Solving a Rational Equation
EXAMPLE
Solve: 3 x  1  6 x  5 .
x4
2x  7
SOLUTION
1) List restrictions on the variable.
3x  1 6 x  5

x  4 2x  7
This denominator would
equal zero if x = 4.
This denominator would
equal zero if x = 3.5.
The restrictions are x  4 and x  3.5.
Blitzer, Intermediate Algebra, 5e – Slide #11 Section 6.6
P 443
Solving a Rational Equation
CONTINUED
2) Multiply both sides by the LCD. The denominators are x – 4
and 2x – 7. Thus, the LCD is (x – 4)(2x - 7).
3x  1 6 x  5

, x  4, x  3.5
x  4 2x  7
This is the given equation.
x  42 x  7 3x  1   x  42 x  7 6 x  5  Multiply both sides by
 x4 
2x  73x  1  x  46x  5
 2 x  7  the LCD.
Simplify.
Blitzer, Intermediate Algebra, 5e – Slide #12 Section 6.6
Solving a Rational Equation
CONTINUED
3) Solve the resulting equation.
2x  73x  1  x  46x  5
This is the equation cleared
of fractions.
6 x 2  19x  7  6 x 2  19x  20
 19 x  7  19 x  20
Use FOIL on each side.
Subtract 6x 2 from both sides.
 7  20
Subtract 19x from both sides.
4) Check the proposed solution in the original equation.
Notice, there is no proposed solution. And of course, -7 = -20 is
not a true statement. Therefore, there is no solution to the original
rational equation. We say the solution set is , the empty set.
Blitzer, Intermediate Algebra, 5e – Slide #13 Section 6.6
Solving a Rational Equation
Do Check Point 2 on page 443
x3 x2

.
x 1 x  6
x  4
The LCD is (x + 1)(x + 6).
The restrictions are x  1 and x  6.
Do Check Point 4 on page 445
x 12
 5
2 x
x  4,6
The LCD is 2x.
The restrictions are x  0
Blitzer, Intermediate Algebra, 5e – Slide #14 Section 6.6
Solving a Rational Equation
EXAMPLE
Solve:
2x 1
2
1


.
2
x  2x  8 x  4 x  2
SOLUTION
1) List restrictions on the variable. By factoring denominators,
it makes it easier to see values that make the denominators zero.
2x 1
2
1


x  4x  2 x  4 x  2
This denominator is zero
if x = -4 or x = 2.
This denominator would
equal zero if x = -4.
The restrictions are x  4 and x  2.
Blitzer, Intermediate Algebra, 5e – Slide #15 Section 6.6
This denominator would
equal zero if x = 2.
Solving a Rational Equation
CONTINUED
2) Multiply both sides by the LCD. The factors of the LCD are
x + 4 and x – 2. Thus, the LCD is (x + 4)(x - 2).
2x 1
2
1


, x  4, x  2
x  4x  2 x  4 x  2
This is the given equation.

2x 1
2 
 1 
  x  4x  2
x  4x  2


 x2
 x  4x  2 x  4 
x  4x  2
Multiply
both sides
by the LCD.
2x 1
2
 1 
 x  4x  2 
 x  4x  2

x  4x  2
x4
 x2
Use the distributive
property.
Blitzer, Intermediate Algebra, 5e – Slide #16 Section 6.6
Solving a Rational Equation
CONTINUED
2 x  1  2x  2  x  4
Simplify.
3) Solve the resulting equation.
2 x  1  2x  2  x  4
2x 1  2x  4  x  4
4x  5  x  4
3x  5  4
3x  9
x3
Blitzer, Intermediate Algebra, 5e – Slide #17 Section 6.6
This is the equation with
cleared fractions.
Use the distributive property.
Combine like terms.
Subtract x from both sides.
Add 5 to both sides.
Divide both sides by 3.
Solving a Rational Equation
CONTINUED
4) Check the proposed solutions in the original equation. The
proposed solution, 3, is not part of the restriction that x  4 and
x  2. Substitute 3 for x, in the given (original) equation. The
resulting true statement verifies that 3 is a solution and that {3} is
the solution set.
Blitzer, Intermediate Algebra, 5e – Slide #18 Section 6.6
Solving a Rational Equation
Do Check Point 5 on page 446
3
5
x 2  20

 2
x  3 x  4 x  7 x  12
The LCD is (x - 3)(x - 4).
The restrictions are
x  4 and x  3
x  1,7
Blitzer, Intermediate Algebra, 5e – Slide #19 Section 6.6
DONE
Solving a Rational Equation
EXAMPLE
Rational functions can be used to model learning. Many of these
functions model the proportion of correct responses as a function
of the number of trials of a particular task. One such model, called
a learning curve, is
0 .9 x  0 .4
f x  
0 . 9 x  0. 1
where f (x) is the proportion of correct responses after x trials. If
f (x) = 0, there are no correct responses. If f (x) = 1, all responses
are correct. The graph of the rational function is shown on the
next page. Use the function to solve the following problem.
Blitzer, Intermediate Algebra, 5e – Slide #21 Section 6.6
Solving a Rational Equation
CONTINUED
Proportion of Correct
Responses
A Learning Curve
1.2
1.1
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
f x  
0
1
2
3
4
5
6
7
8
0 .9 x  0 .4
0 . 9 x  0. 1
9 10 11 12 13 14 15
Number of Learning Trials
Blitzer, Intermediate Algebra, 5e – Slide #22 Section 6.6
Solving a Rational Equation
CONTINUED
How many learning trials are necessary for 0.5 of the responses to
be correct? Identify your solution as a point on the graph.
SOLUTION
Substitute 0.5, the proportion of correct responses, for f (x) and
solve the resulting rational equation for x.
0 .9 x  0 .4
0 .5 
0 .9 x  0 .1
0.9 x  0.4
0.9 x  0.1  0.5  0.9 x  0.1 
0.9 x  0.1
0.50.9 x  0.1  0.9 x  0.4
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 6.6
The LCD is 0.9x + 0.1.
Multiply both sides by the
LCD.
Simplify.
Solving a Rational Equation
CONTINUED
0.45 x  0.05  0.9 x  0.4
 0.45 x  0.05  0.4
 0.45 x  0.45
x 1
Use the distributive property
on the left side.
Subtract 0.9x from both sides.
Subtract 0.05 from both sides.
Divide both sides by -0.45.
The number of learning trials necessary for 0.5 of the responses to
be correct is 1.
The solution is identified as a point on the graph at the beginning of
the problem.
Blitzer, Intermediate Algebra, 5e – Slide #24 Section 6.6
Solving a Rational Equation
Important to Remember:
A common error when solving rational equations is to forget to list
the restrictions for the variable in the very beginning and then to
reject any of those values in the end as solutions.
Don’t forget to say in the beginning what x can’t be! And don’t forget to
throw out a solution if it’s something x “can’t be”….
Another common error is to check the proposed solutions in one
of the later equations.
You must check your proposed solutions in the original equation.
Blitzer, Intermediate Algebra, 5e – Slide #25 Section 6.6
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