Solving
Rational
Equations
Solving
Rational
Equations
5-5
5-5 and Inequalities
and Inequalities
Warm Up
Lesson Presentation
Lesson Quiz
HoltMcDougal
Algebra 2Algebra 2
Holt
5-5
Solving Rational Equations
and Inequalities
Warm Up
Add or subtract. Identify any x-values for
which the expression is undefined.
1.
2.
1
4x
+
1
4
–
x
x
3.
Holt McDougal Algebra 2
1
4x
1
2x
3
x
x≠0
5-5
Solving Rational Equations
and Inequalities
Objective
Solve rational equations and
inequalities.
Holt McDougal Algebra 2
5-5
Solving Rational Equations
and Inequalities
Essential Question
What are the steps for solving rational
equations and inequalities?
Holt McDougal Algebra 2
5-5
Solving Rational Equations
and Inequalities
Vocabulary
rational equation
extraneous solution
rational inequality
Holt McDougal Algebra 2
5-5
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 McDougal Algebra 2
5-5
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 McDougal Algebra 2
5-5
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 McDougal Algebra 2
Solve for x.
5-5
Solving Rational Equations
and Inequalities
Example 1 Continued
18
Check x – x = 3
6 – 18
6
6–3
3
Holt McDougal Algebra 2
3
3
3 
x – 18 = 3
x
18
(–3) –(–3)
–3 + 6
3
3
3
3 
5-5
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 McDougal Algebra 2
Multiply each term by
the LCD, 3x.
Simplify. Note that x ≠ 0.
Combine like terms.
Solve for x.
5-5
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 McDougal Algebra 2
Multiply each term by
the LCD, 4x.
Simplify. Note that x ≠ 0.
Combine like terms.
Solve for x.
5-5
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 McDougal Algebra 2
Solve for x.
5-5
Solving Rational Equations
and Inequalities
Essential Question
What are the steps for solving rational
equations?
1.
2.
3.
4.
5.
6.
Multiply each term by the LCD.
Simplify.
Write in standard form by combining like terms.
Factor the equation if needed.
Set each factor or equation equal to 0.
Solve each equation.
Holt McDougal Algebra 2
5-5
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 McDougal Algebra 2
5-5
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 (x – 2) the LCD, x – 2.
5x
3x + 4
Divide out common
(x
–
2)
=
(x
–
2)
x–2
x–2
factors.
5x = 3x + 4
x=2
Simplify. Note that x ≠ 2.
Solve for x.
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 McDougal Algebra 2
5-5
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 McDougal Algebra 2
Division by 0 is
undefined.
5-5
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 McDougal Algebra 2
5-5
Solving Rational Equations
and Inequalities
You can 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 McDougal Algebra 2
5-5
Solving Rational Equations
and Inequalities
Example 6: Solving Rational Inequalities
Algebraically
Solve
6
x– 8
≤ 3 algebraically.
Step 1 Solve for x.
6 (x – 8) = 3(x – 8)
x–8
6 = 3x – 24
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.
Holt McDougal Algebra 2
5-5
Solving Rational Equations
and Inequalities
Example 6 Continued
Solve x – 8
= 0 algebraically.
x–8=0
x=8
Solve for x.
x ≤ 8 or x ≥ 10
The boundary points are 10 and 8. Graph on
a number line. Test points to the left of 8,
between 8 and 10, and to the right of 10.
Which satisfies the original inequality?
6
≤3
x– 8
8
10
Holt McDougal Algebra 2
5-5
Solving Rational Equations
and Inequalities
Check It Out! Example 6a
Solve
6
x– 2
≥ –4 algebraically.
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 McDougal Algebra 2
Multiply by the LCD.
Simplify. Note that x ≠ 2.
Solve for x.
Rewrite with the variable
on the left.
5-5
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 McDougal Algebra 2
≥ –4 is x >
5-5
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 McDougal Algebra 2
5-5
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 McDougal Algebra 2
5-5
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 McDougal Algebra 2
5-5
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 McDougal Algebra 2
5-5
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 McDougal Algebra 2
5-5
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 McDougal Algebra 2
5-5
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 McDougal Algebra 2
5-5
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
Combine like terms.
–5c2 = –12
Solve for c.
c ≈ ± 1.5
The speed of the current cannot be negative. Therefore,
the average speed of the current is about 1.5 mi/h.
Holt McDougal Algebra 2
5-5
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 McDougal Algebra 2