Transcript ppsx

Solving
Rational
Equations
Solving
Rational
Equations
8-5
8-5 and Inequalities
and Inequalities
Warm Up
Lesson Presentation
Lesson Quiz
Holt
Holt
Algebra
Algebra
22
8-5
Solving Rational Equations
and Inequalities
Warm Up
Find the least common multiple for each
pair.
1. 2x2 and 4x2 – 2x
2x2(2x – 1)
2. x + 5 and x2 – x – 30
(x + 5)(x – 6)
Add or subtract. Identify any x-values for
which the expression is undefined.
1
1
3.
5x – 2
+
x–2
4x
4x(x – 2)
4.
1
1
–
x
x2
Holt Algebra 2
–(x – 1)
x2
x≠0
8-5
Solving Rational Equations
and Inequalities
Objective
Solve rational equations and
inequalities.
Holt Algebra 2
8-5
Solving Rational Equations
and Inequalities
Vocabulary
rational equation
extraneous solution
rational inequality
Holt Algebra 2
8-5
Solving Rational Equations
and Inequalities
A rational equation is an equation that contains
one or more rational expressions. The time t in hours
that it takes to travel d miles can be determined by
using the equation t = d , where r is the average rate
r
of speed. This equation is a rational equation.
Holt Algebra 2
8-5
Solving Rational Equations
and Inequalities
To solve a rational equation, start by multiplying
each term of the equation by the least common
denominator (LCD) of all of the expressions in the
equation. This step eliminates the denominators of
the rational expression and results in an equation
you can solve by using algebra.
Holt Algebra 2
8-5
Solving Rational Equations
and Inequalities
Example 1: Solving Rational Equations
Solve the equation x – 18 = 3.
x
x(x) – 18 (x) = 3(x) Multiply each term by the LCD, x.
x
Simplify. Note that x ≠ 0.
x2 – 18 = 3x
x2 – 3x – 18 = 0
(x – 6)(x + 3) = 0
Write in standard form.
Factor.
x – 6 = 0 or x + 3 = 0 Apply the Zero Product Property.
x = 6 or x = –3
Holt Algebra 2
Solve for x.
8-5
Solving Rational Equations
and Inequalities
Example 1 Continued
=3
Check x – 18
x
6 – 18
6
6–3
3
Holt Algebra 2
3
3
3 
x – 18 = 3
x
18
(–3) –(–3)
–3 + 6
3
3
3
3 
8-5
Solving Rational Equations
and Inequalities
Check It Out! Example 1a
Solve the equation 10 = 4 + 2.
x
3
10 (3x) = 4 (3x) + 2(3x)
x
3
10x = 12 + 6x
4x = 12
x=3
Holt Algebra 2
Multiply each term by
the LCD, 3x.
Simplify. Note that x ≠ 0.
Combine like terms.
Solve for x.
8-5
Solving Rational Equations
and Inequalities
Check It Out! Example 1b
Solve the equation 6 + 5 = – 7 .
4
x
4
6 (4x) + 5 (4x) = – 7 (4x)
4
4
x
24 + 5x = –7x
24 = –12x
x = –2
Holt Algebra 2
Multiply each term by
the LCD, 4x.
Simplify. Note that x ≠ 0.
Combine like terms.
Solve for x.
8-5
Solving Rational Equations
and Inequalities
Check It Out! Example 1c
Solve the equation x = 6 – 1.
x
x(x) = 6 (x) – 1(x) Multiply each term by the LCD, x.
x
Simplify. Note that x ≠ 0.
x2 = 6 – x
x2 + x – 6 = 0
(x – 2)(x + 3) = 0
Write in standard form.
Factor.
x – 2 = 0 or x + 3 = 0 Apply the Zero Product Property.
x = 2 or x = –3
Holt Algebra 2
Solve for x.
8-5
Solving Rational Equations
and Inequalities
An extraneous solution is a solution of an
equation derived from an original equation that
is not a solution of the original equation. When
you solve a rational equation, it is possible to get
extraneous solutions. These values should be
eliminated from the solution set. Always check
your solutions by substituting them into the
original equation.
Holt Algebra 2
8-5
Solving Rational Equations
and Inequalities
Example 2A: Extraneous Solutions
Solve each equation.
5x = 3x + 4
x–2
x–2
Multiply each term by
5x
3x
+
4
(x – 2) =
(x – 2)
x–2
the LCD, x – 2.
x–2
5x
3x + 4 (x – 2) Divide out common
(x
–
2)
=
x–2
x–2
factors.
Simplify. Note that x ≠ 2.
5x = 3x + 4
Solve for x.
x=2
The solution x = 2 is extraneous because it makes
the denominators of the original equation equal to 0.
Therefore, the equation has no solution.
Holt Algebra 2
8-5
Solving Rational Equations
and Inequalities
Example 2A Continued
Check Substitute 2 for x in the original equation.
5x
3x + 4
=
x–2
x–2
5(2)
3(2) + 4
2–2
2–2
10
10
0
0
Holt Algebra 2
Division by 0 is
undefined.
8-5
Solving Rational Equations
and Inequalities
Example 2B: Extraneous Solutions
Solve each equation.
2x – 5 + x =
11
x–8
x–8
2
Multiply each term by the LCD, 2(x – 8).
2x – 5
x 2(x – 8) =
11
2(x
–
8)
+
2(x – 8)
x–8
x–8
2
Divide out common factors.
2x – 5
x 2(x – 8) =
11
2(x
–
8)
+
2(x – 8)
x–8
x–8
2
2(2x – 5) + x(x – 8) = 11(2)
4x – 10 + x2 – 8x = 22
Holt Algebra 2
Simplify. Note that x ≠ 8.
Use the Distributive
Property.
8-5
Solving Rational Equations
and Inequalities
Example 2B Continued
x2 – 4x – 32 = 0
(x – 8)(x + 4) = 0
Write in standard form.
Factor.
x – 8 = 0 or x + 4 = 0 Apply the Zero Product Property.
x = 8 or x = –4
Solve for x.
The solution x = 8 us extraneous because it
makes the denominator of the original equation
equal to 0. The only solution is x = –4.
Holt Algebra 2
8-5
Solving Rational Equations
and Inequalities
Check
Example 2B Continued
Write
2x – 5
x
11
+
=
x–8
x–8
2
as
Graph the left side of the
equation as Y1. Identify
the values of x for which
Y1 = 0.
The graph intersects the
x-axis only when x = –4.
Therefore, x = –4 is the
only solution.
Holt Algebra 2
2x – 5
x
11
+
–
= 0.
x–8
x
–
8
2
8-5
Solving Rational Equations
and Inequalities
Check It Out! Example 2a
16
2
Solve the equation
.
=
2
x – 16
x–4
Multiply each term by the LCD, (x – 4)(x +4).
16
(x – 4)(x + 4) (x – 4)(x + 4) =
Divide out common factors.
16
(x – 4)(x + 4) (x – 4)(x + 4) =
2
(x – 4 )(x + 4)
x–4
2
(x – 4 )(x + 4)
x–4
Simplify. Note that x ≠ ±4.
16 = 2x + 8
Solve for x.
x=4
The solution x = 4 is extraneous because it makes
the denominators of the original equation equal to 0.
Therefore, the equation has no solution.
Holt Algebra 2
8-5
Solving Rational Equations
and Inequalities
Check It Out! Example 2b
1
x
x .
=
+
x–1
x–1
6
Multiply each term by the LCD, 6(x – 1).
1
x
x 6(x – 1)
6(x
–
1)
=
6(x
–
1)
+
x–1
x–1
6
Solve the equation
Divide out common factors.
1
x
x 6(x – 1)
6(x
–
1)
=
6(x
–
1)
+
x–1
x–1
6
6 = 6x + x(x – x) Simplify. Note that x ≠ 1.
Use the Distributive
6 = 6x + x2 – x
Property.
Holt Algebra 2
8-5
Solving Rational Equations
and Inequalities
Check It Out! Example 2b Continued
0 = x2 + 5x – 6
Write in standard form.
0 = (x + 6)(x – 1) Factor.
x + 6 = 0 or x – 1 = 0 Apply the Zero Product Property.
x = –6 or x = 1
Solve for x.
The solution x = 1 us extraneous because it
makes the denominator of the original equation
equal to 0. The only solution is x = –6.
Holt Algebra 2
8-5
Solving Rational Equations
and Inequalities
Example 3: Problem-Solving Application
A jet travels 3950 mi from Chicago, Illinois,
to London, England, and 3950 mi on the
return trip. The total flying time is 16.5 h.
The return trip takes longer due to winds
that generally blow from west to east. If the
jet’s average speed with no wind is 485
mi/h, what is the average speed of the wind
during the round-trip flight? Round to the
nearest mile per hour.
Holt Algebra 2
8-5
Solving Rational Equations
and Inequalities
Example 3 Continued
1
Understand the Problem
The answer will be the average speed of
the wind.
List the important information:
• The jet spent 16.5 h on the round-trip.
• It went 3950 mi east and 3950 mi west.
• Its average speed with no wind is 485 mi/h.
Holt Algebra 2
8-5
Solving Rational Equations
and Inequalities
Example 3 Continued
2
Make a Plan
Let w represent
Average
the speed of the
Distance
Speed
wind. When the jet
(mi)
(mi/h)
Time (h)
3950
is going east, its
East
3950
485 + w 485 + w
speed is equal to
3950
West
3950
485 – w
its speed with no
485 – w
wind plus w. When
the jet is going
total time = time east + time west
west, its speed is
3950 + 3950
equal to its speed
16.5 = 485
+w
485 – w
with no wind
minus w.
Holt Algebra 2
8-5
3
Solving Rational Equations
and Inequalities
Solve
The LCD is (485 + w)(485 – w).
16.5(485 + w)(485 – w) =
+
3950
485 – w
3950
485 + w
(485 + w)(485 – w)
(485 + w)(485 – w)
Simplify. Note that x ≠ ±485.
16.5(485 + w)(485 – w) = 3950(485 – w) + 3950 (485 + w)
Use the Distributive Property.
3,881,212.5 – 16.5w2 = 1,915,750 – 3950w + 1,915,750 + 3950w
3,881,212.5 – 16.5w2 = 3,831,500 Combine like terms.
–16.5w2 = –49,712.5 Solve for w.
w ≈ ± 55
The speed of the wind cannot be negative. Therefore,
the average speed of the wind is 55 mi/h.
Holt Algebra 2
8-5
Solving Rational Equations
and Inequalities
Example 3 Continued
4
Look Back
If the speed of the wind is 55 mi/h, the jet’s speed
when going east is 485 + 55 = 540 mi/h. It will
take the jet approximately 7.3 h to travel 3950 mi
east. The jet’s speed when going west is 485 – 55
= 430 mi/h. It will take the jet approximately 9.2
h to travel 3950 mi west. The total trip will take
16.5 h, which is the given time.
Holt Algebra 2
8-5
Solving Rational Equations
and Inequalities
Check It Out! Example 3
On a river, a kayaker travels 2 mi upstream
and 2 mi downstream in a total of 5 h. In
still water, the kayaker can travel at an
average speed of 2 mi/h. Based on this
information, what is the average speed of
the current of this river? Round to the
nearest tenth.
Holt Algebra 2
8-5
Solving Rational Equations
and Inequalities
Check It Out! Example 3 Continued
1
Understand the Problem
The answer will be the average speed of
the current.
List the important information:
• The kayaker spent 5 hours kayaking.
• She went 2 mi upstream and 2 mi
downstream.
• Her average speed in still water is 2 mi/h.
Holt Algebra 2
8-5
Solving Rational Equations
and Inequalities
Check It Out! Example 3 Continued
2
Make a Plan
Let c represent the
Average
speed of the
Distance
Speed
current. When the
(mi)
(mi/h)
Time (h)
kayaker is going
2
Up
2
2
–
c
upstream, her speed
2–c
is equal to her
2
Down
2
2+c
2+c
speed in still water
minus c. When the
total time = time up- + time downkayaker is going
stream
stream
downstream, her
speed is equal to
2
2
5
=
+
her speed in still
2–c
2+c
water plus c.
Holt Algebra 2
8-5
3
Solving Rational Equations
and Inequalities
Solve
5(2 + c)(2 – c) =
The LCD is (2 – c)(2 + c).
2
2
(2 + c)(2 – c) +
(2 + c)(2 – c)
2–c
2+c
Simplify. Note that x ≠ ±2.
5(2 + c)(2 – c) = 2(2 + c) + 2(2 – c)
Use the Distributive Property.
20 – 5c2 = 4 + 2c + 4 – 2c
20 – 5c2 = 8
–5c2 = –12
c ≈ ± 1.5
Combine like terms.
Solve for c.
The speed of the current cannot be negative. Therefore,
the average speed of the current is about 1.5 mi/h.
Holt Algebra 2
8-5
Solving Rational Equations
and Inequalities
Check It Out! Example 3 Continued
4
Look Back
If the speed of the current is about 1.5 mi/h, the
kayaker’s speed when going upstream is 2 – 1.5 =
0.5 mi/h. It will take her about 4 h to travel 2 mi
upstream. Her speed when going downstream is
about 2 + 1.5 = 3.5 mi/h. It will take her 0.5 h to
travel 2 mi downstream. The total trip will take
about 4.5 hours which is close to the given time of
5 h.
Holt Algebra 2
8-5
Solving Rational Equations
and Inequalities
Example 4: Work Application
Natalie can finish a 500-piece puzzle in about
8 hours. When Natalie and Renzo work
together, they can finish a 500-piece puzzle in
about 4.5 hours. About how long will it take
Renzo to finish a 500-piece puzzle if he works
by himself?
Natalie’s rate: 1 of the puzzle per hour
8
Renzo’s rate: 1 of the puzzle per hour, where h is the
h
number of hours needed to finish the puzzle by himself.
Holt Algebra 2
8-5
Solving Rational Equations
and Inequalities
Example 4 Continued
Natalie’s rate
+ Renzo’s rate
 hours worked
 hours worked
1 (4.5)
8
+
1 (4.5)
h
= 1 complete
puzzle
=
1
1 (4.5)(8h) + 1 (4.5)(8h) = 1(8h) Multiply by the LCD,8h.
8
h
4.5h + 36 = 8h
Simplify.
36 = 3.5h
10.3 = h
Solve for h.
It will take Renzo about 10.3 hours, or 10 hours 17 minutes
to complete a 500-piece puzzle working by himself.
Holt Algebra 2
8-5
Solving Rational Equations
and Inequalities
Check It Out! Example 4
Julien can mulch a garden in 20 minutes.
Together Julien and Remy can mulch the
same garden in 11 minutes. How long will it
take Remy to mulch the garden when
working alone?
Julien’s rate: 1 of the garden per minute
20
Remy’s rate: 1 of the garden per minute, where m is
m
the number of minutes needed to mulch the garden by
himself.
Holt Algebra 2
8-5
Solving Rational Equations
and Inequalities
Check It Out! Example 4 Continued
Julien’s rate
 min worked
1 (11)
20
+ Remy’s rate
 min worked
+
1 (11)
m
= 1 complete
job
=
1
1 (11)(20m)+ 1 (11)(20m) = 1(20m) Multiply by the LCD,
20m.
20
m
11m + 220 = 20m
Simplify.
220 = 9m
24 ≈ m
Solve for m.
It will take Remy about 24 minutes to mulch the garden
working by himself.
Holt Algebra 2
8-5
Solving Rational Equations
and Inequalities
A rational inequality is an inequality that
contains one or more rational expressions. One
way to solve rational inequalities is by using
graphs and tables.
Holt Algebra 2
8-5
Solving Rational Equations
and Inequalities
Example 5: Using Graphs and Tables to Solve
Rational Equations and Inequalities
Solve
x
x– 6
≤ 3 by using a graph and a table.
Use a graph. On a
graphing calculator,
Y1 = x
and Y2 = 3.
x– 6
The graph of Y1 is at
or below the graph of
Y2 when x < 6 or
when x ≥ 9.
Holt Algebra 2
(9, 3)
Vertical
asymptote:
x=6
8-5
Solving Rational Equations
and Inequalities
Example 5 Continued
Use a table. The table shows that Y1 is
undefined when x = 6 and that Y1 ≤ Y2
when x ≥ 9.
The solution of the inequality is x < 6 or x ≥ 9.
Holt Algebra 2
8-5
Solving Rational Equations
and Inequalities
Check It Out! Example 5a
Solve
x
x– 3
≥ 4 by using a graph and a table.
Use a graph. On a
graphing calculator,
Y1 = x
and Y2 = 4.
x– 3
The graph of Y1 is at
or below the graph of
Y2 when x < 3 or
when x ≥ 4.
Holt Algebra 2
(4, 4)
Vertical
asymptote:
x=3
8-5
Solving Rational Equations
and Inequalities
Check It Out! Example 5a continued
Use a table. The table shows that Y1 is
undefined when x = 3 and that Y1 ≤ Y2
when x ≥ 4.
The solution of the inequality is x < 3 or x ≥ 4.
Holt Algebra 2
8-5
Solving Rational Equations
and Inequalities
Check It Out! Example 5b
Solve
8
x+ 1
= –2 by using a graph and a table.
Use a graph. On a
graphing calculator,
8 and Y2 = –2.
Y1 =
x+ 1
The graph of Y1 is at
or below the graph of
Y2 when x = –5.
Holt Algebra 2
(–5, –2)
Vertical
asymptote:
x = –1
8-5
Solving Rational Equations
and Inequalities
Check It Out! Example 5b continued
Use a table. The table shows that Y1 is
undefined when x = –1 and that Y1 ≤ Y2
when x = –5.
The solution of the inequality is x = –5.
Holt Algebra 2
8-5
Solving Rational Equations
and Inequalities
You can also solve rational inequalities
algebraically. You start by multiplying each
term by the least common denominator
(LCD) of all the expressions in the inequality.
However, you must consider two cases: the
LCD is positive or the LCD is negative.
Holt Algebra 2
8-5
Solving Rational Equations
and Inequalities
Example 6: Solving Rational Inequalities
Algebraically
Solve
6
x– 8
≤ 3 algebraically.
Case 1 LCD is positive.
Step 1 Solve for x.
6 (x – 8) ≤ 3(x – 8)
x–8
6 ≤ 3x – 24
Holt Algebra 2
Multiply by the LCD.
Simplify. Note that x ≠ 8.
30 ≤ 3x
Solve for x.
10 ≤ x
x ≥ 10
Rewrite with the variable
on the left.
8-5
Solving Rational Equations
and Inequalities
Example 6 Continued
Solve
6
x– 8
≤ 3 algebraically.
Step 2 Consider the sign of the LCD.
x–8>0
LCD is positive.
x>8
Solve for x.
For Case 1, the solution must satisfy x ≥ 10
and x > 8, which simplifies to x ≥ 10.
Holt Algebra 2
8-5
Solving Rational Equations
and Inequalities
Example 6: Solving Rational Inequalities
Algebraically
Solve
6
x– 8
≤ 3 algebraically.
Case 2 LCD is negative.
Step 1 Solve for x.
6 (x – 8) ≥ 3(x – 8)
x–8
6 ≥ 3x – 24
Holt Algebra 2
Multiply by the LCD.
Reverse the inequality.
Simplify. Note that x ≠ 8.
30 ≥ 3x
Solve for x.
10 ≥ x
x ≤ 10
Rewrite with the variable
on the left.
8-5
Solving Rational Equations
and Inequalities
Example 6 Continued
Solve
6
x– 8
≤ 3 algebraically.
Step 2 Consider the sign of the LCD.
x–8>0
LCD is positive.
Solve for x.
x>8
For Case 2, the solution must satisfy x ≤ 10
and x < 8, which simplifies to x < 8.
The solution set of the original inequality is the
union of the solutions to both Case 1 and Case 2.
The solution to the inequality
6
x– 8
or x ≥ 10, or {x|x < 8  x ≥ 10}.
Holt Algebra 2
≤ 3 is x < 8
8-5
Solving Rational Equations
and Inequalities
Check It Out! Example 6a
Solve
6
x– 2
≥ –4 algebraically.
Case 1 LCD is positive.
Step 1 Solve for x.
6 (x – 2) ≥ –4(x – 2)
x–2
6 ≥ –4x + 8
–2 ≥ –4x
1
≤x
2
1
x≥
2
Holt Algebra 2
Multiply by the LCD.
Simplify. Note that x ≠ 2.
Solve for x.
Rewrite with the variable
on the left.
8-5
Solving Rational Equations
and Inequalities
Check It Out! Example 6a Continued
Solve
6
x– 2
≥ –4 algebraically.
Step 2 Consider the sign of the LCD.
LCD is positive.
x–2>0
x>2
Solve for x.
1
For Case 1, the solution must satisfy x ≥
2
and x > 2, which simplifies to x > 2.
Holt Algebra 2
8-5
Solving Rational Equations
and Inequalities
Check It Out! Example 6a Continued
Solve
6
x– 2
≥ –4 algebraically.
Case 2 LCD is negative.
Step 1 Solve for x.
6 (x – 2) ≤ –4(x – 2)
x–2
6 ≤ –4x + 8
–2 ≤ –4x
1
≥x
2
1
x≤
2
Holt Algebra 2
Multiply by the LCD.
Reverse the inequality.
Simplify. Note that x ≠ 2.
Solve for x.
Rewrite with the variable
on the left.
8-5
Solving Rational Equations
and Inequalities
Check It Out! Example 6a Continued
6
Solve
≥ –4 algebraically.
x– 2
Step 2 Consider the sign of the LCD.
LCD is negative.
x–2<0
Solve for x.
x<2
1
For Case 2, the solution must satisfy x ≤
2
1
and x < 2, which simplifies to x ≤ .
2
The solution set of the original inequality is the
union of the solutions to both Case 1 and Case 2.
6
The solution to the inequality
x– 2
1
1
2 or x ≤
, or {x| x ≤
 x > 2}.
2
2
Holt Algebra 2
≥ –4 is x >
8-5
Solving Rational Equations
and Inequalities
Check It Out! Example 6b
Solve
9
x+ 3
< 6 algebraically.
Case 1 LCD is positive.
Step 1 Solve for x.
9 (x + 3) < 6(x + 3)
x+3
9 < 6x + 18
–9 < 6x
3
– <x
2
3
x>–
2
Holt Algebra 2
Multiply by the LCD.
Simplify. Note that x ≠ –3.
Solve for x.
Rewrite with the variable
on the left.
8-5
Solving Rational Equations
and Inequalities
Check It Out! Example 6b Continued
Solve
9
x+ 3
< 6 algebraically.
Step 2 Consider the sign of the LCD.
LCD is positive.
x+3>0
x > –3
Solve for x.
For Case 1, the solution must satisfy x >– 3
2
3
and x > –3, which simplifies to x >–
.
2
Holt Algebra 2
8-5
Solving Rational Equations
and Inequalities
Check It Out! Example 6b Continued
Solve
9
x+ 3
< 6 algebraically.
Case 2 LCD is negative.
Step 1 Solve for x.
9 (x + 3) > 6(x + 3)
x+3
9 > 6x + 18
–9 > 6x
3
– >x
2
3
x<–
2
Holt Algebra 2
Multiply by the LCD.
Reverse the inequality.
Simplify. Note that x ≠ –3.
Solve for x.
Rewrite with the variable
on the left.
8-5
Solving Rational Equations
and Inequalities
Check It Out! Example 6b Continued
9
Solve
< 6 algebraically.
x+ 3
Step 2 Consider the sign of the LCD.
LCD is negative.
x+3<0
Solve for x.
x < –3
3
For Case 2, the solution must satisfy x <–
2
and x < –3, which simplifies to x < –3.
The solution set of the original inequality is the
union of the solutions to both Case 1 and Case 2.
9
The solution to the inequality
< 6 is x < –3
x+ 3
3
3
or x > –
, or {x| x > –
 x < –3}.
2
2
Holt Algebra 2
8-5
Solving Rational Equations
and Inequalities
Lesson Quiz
Solve each equation or inequality.
1.
2.
x+2 =
x
x–1
2
6x
7x + 4
=
x+4
x+4
x = –1 or x = 4
no solution
x+2
5
x = –5
+ x =
x–3
x–3
5
4
4.
≥2
3<x≤5
x– 3
5. A college basketball player has made 58 out of 82
attempted free-throws this season. How many
additional free-throws must she make in a row to
raise her free-throw percentage to 90%? 158
3.
Holt Algebra 2