Download document

1
Equations and
Inequalities
Sections 1.1–1.4
© 2008 Pearson Addison-Wesley.
All rights reserved
1 Equations and Inequalities
1.1 Linear Equations
1.2 Applications and Modeling with
Linear Equations
1.1 Linear Equations
Basic Terminology of Equations ▪ Solving Linear Equations ▪
Identities, Conditional Equations, and Contradictions ▪ Solving
for a Specified Variable (Literal Equations)
Basic Terminology of Equations
An equation is a statement that two expressions are equal.
2
x

2

9
,
11
x

5
x

6
x
,
x
 2x  1  0
For example:
To solve an equation means to find all numbers that make
the equation a true statement. These numbers are called the
solutions or roots of the equation. These solutions make up
the solution set. Equations with the same solution set are
called equivalent equations.
We will focus on solving linear equations. A linear equation in
one variable is an equation that can be written in the form
ax + b = 0
where a and b are real numbers and a ≠ 0.
Addition and Multiplication
Properties of Equality
For real numbers a, b, and c:
If a = b, then a + c = b + c.
That is, the same number or expression may be added to
(or subtracted from) both sides of an equation without
changing the solution set.
If a = b and c ≠ 0, then ac = bc.
That is, both sides of an equation may be multiplied (or
divided) by the same nonzero number without changing
the solution set.
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
1-5
1.1
Example 1 Solving a Linear Equation
Solve
(page 85)
.
Distributive property
Combine terms.
Add 4 to both sides.
Add 12x to both sides.
Combine terms.
Divide both sides by 4.
Solution set: {6}
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
1-6
1.1
Example 2 Clearing Fractions Before Solving a Linear
Equation (page 85)
Solve
.
Multiply by 10, the LCD of
all the fractions.
Distributive property
Combine terms.
Add –4s and –6 to both
sides. Combine terms.
Divide both sides by –6.
Solution set: {–10}
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
1-7
Types of Equations
1. Identity – is an equation satisfied by every
number.
2. Conditional Equations – is an equation satisfied
by some numbers but not others.
3. Contradiction – is an equation that has no
solution.
Indentifying the type of linear equation comes from
whether the resulting statement is true (1),
unique (2) or false (3).
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
1-8
1.1
Example 3(a) Identifying Types of Equations (page 86)
Decide whether the equation is an identity, a
conditional equation, or a contradiction. Give the
solution set.
Add –4x and 9 to both
sides. Combine terms.
Divide both sides by 2.
This is a conditional equation.
Solution set: {11}
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
1-9
1.1
Example 3(b) Identifying Types of Equations (page 86)
Decide whether the equation is an identity, a
conditional equation, or a contradiction. Give the
solution set.
Distributive property
Subtract 14x from both
sides.
This is a contradiction.
Solution set: ø
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
1-10
1.1
Example 3(c) Identifying Types of Equations (page 86)
Decide whether the equation is an identity, a
conditional equation, or a contradiction. Give the
solution set.
Distributive property
Combine terms.
Add x and –3 to both
sides.
This is an identity.
Solution set: {all real numbers}
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
1-11
1.1
Example 4 Solving for a Specified Variable
(page 87)
Solve for the specified variable.
(a) d = rt, for t
Divide both sides by r.
(b)
, for k
Factor out k.
Divide both sides by
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
1-12
1.1
Example 5 Applying the Simple Interest Formula
(page 88)
Caden borrowed $2580 to buy new kitchen
appliances for his home. He will pay off the loan in 9
months at an annual simple interest rate of 6.0%.
How much interest will he pay?
Caden will pay $116.10 in interest.
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
1-13
1.2
Applications and Modeling with Linear
Equations
Solving Applied Problems ▪ Geometry Problems ▪ Motion
Problems ▪ Mixture Problems ▪ Modeling with Linear Equations
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
1-14
Step for Solving an Applied Problem
1. Read the problem carefully until you understand
what is given and what is to be found.
2. Assign a variable to represent the unknown
value, using diagrams or tables as needed.
3. Write an equation using the variable
expressions.
4. Solve the equation.
5. State the answer to the problem.
6. Check the answer in the words of the original
problem.
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
1-15
1.2
Example 1 Find the Dimensions of a Square
(page 91)
The length of a rectangle is 2 in. more than the width.
If the length and width are each increased by 3 in.,
the perimeter of the new rectangle will be 4 in. less
than 8 times the width of the original rectangle. Find
the dimensions of the original rectangle.
Assign variables:
Let x = the length of the original rectangle.
Then, x − 2 = the width of the original rectangle.
x + 3 = the length of the new rectangle
(x − 2) + 3 = x + 1 = the width of the new rectangle.
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
1-16
1.2
Example 1 Find the Dimensions of a Square
(cont.)
The perimeter of the new rectangle is
The perimeter of the new rectangle is 4 in. less than
8 times the width of the original rectangle, so we
have
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
1-17
1.2
Example 1 Find the Dimensions of a Square
(cont.)
Distributive property.
Combine terms.
Add –4x and 20 to
both sides.
Divide both sides by 4.
The length of the original rectangle is 7 in.
The width of the original rectangle is 7 – 2 = 5 in.
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
1-18
1.2
Example 2 Solving a Motion Problem
(page 92)
Krissa drove to her grandmother’s house. She
averaged 40 mph driving there. She was able to
average 48 mph returning, and her driving time
was 1 hr less. What is the distance between
Krissa’s house and her grandmother’s house?
Let x = the distance between Krissa’s house and
her grandmother’s house
Use d = rt
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
1-19
1.2
Example 2 Solving a Motion Problem
(cont.)
The driving time returning was 1 hour less than the
driving time going, so
Multiply by 40·48.
Distributive property.
Subtract 48x.
Divide by –8.
The distance from Krissa’s home to her
grandmother’s home is 240 mi.
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
1-20
1.2
Example 3 Solving a Mixture Problem
(page 93)
How many liters of a 25% anti-freeze solution
should be added to 5 L of a 10% solution to obtain a
15% solution?
Let x = the amount of 25% solution
The number of liters of pure antifreeze in the 25%
solution plus the number of liters of pure
antifreeze in the 10% solution must equal the
number of liters of pure antifreeze in the 15%
solution.
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
1-21
1.2
Example 3 Solving a Mixture Problem
(cont.)
Create a table to show the relationships in the problem.
Write an equation:
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
1-22
1.2
Example 3 Solving a Mixture Problem
(cont.)
Distributive property.
Subtract .15x and .5.
Divide by .1.
2.5 liters of the 25% solution should be added.
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
1-23
1.2
Example 4 Solving an Investment Problem
(page 94)
Last year, Owen earned a total of $1456 in interest
from two investments. He invested a total of
$28,000, part at 4.8% and the rest at 5.5%. How
much did he invest at each rate?
Let x = amount invested at 4.8%.
Then 28,000 − x = amount invested at 5.5%.
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
1-24
1.2
Example 4 Solving an Investment Problem
(cont.)
Create a table to show the relationships in the problem.
The amount of interest from the 4.8% account
plus the amount of interest from the 5.5%
account must equal the total amount of
interest.
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
1-25
1.2
Example 4 Solving an Investment Problem
(cont.)
Distributive property
Combine terms.
Subtract 1540.
Divide by –.007.
Owen invested $12,000 at 4.8% and
$28,000 − $12,000 = $16,000 at 5.5%.
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
1-26
1.2
Example 5 Modeling the Prevention of Indoor Pollutants
(page 95)
A range hood removes contaminants at a rate of F
liters of air per second. The percent P of contaminants
that are also removed from the surrounding air can be
modeled by the linear equation
where 10 ≤ F ≤ 75. What flow F must a range hood
have to remove 70% of the contaminants from the air?
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
1-27
1.2
Example 5 Modeling the Prevention of Indoor Pollutants
(cont.)
Substitute P = 70 into
for F.
and solve
Subtract 7.18.
Divide by 1.06.
The flow rate must be approximately 59.26 L per
second to remove 70% of the contaminants.
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
1-28