Transcript Solving Basic Linear Equations in One Variable

Equations and Inequalities
Copyright © Cengage Learning. All rights reserved.
2
Section
2.1
Solving Basic Linear Equations
in One Variable
Copyright © Cengage Learning. All rights reserved.
Objectives
1. Determine whether a statement is an
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expression or an equation.
22. Determine whether a number is a solution of
an equation.
33. Solve a linear equation in one variable by
applying the addition or subtraction property of
equality.
4. Solve a linear equation in one variable by
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applying the multiplication or division property of
equality.
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Objectives
5 Application 1.
6 Application 2
7 5. Application 3
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1.
Determine whether a statement is an
expression or an equation
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Expression or Equation
Equation: a statement indicating that two quantities are
equal.
•
•
•
x + 5 = 21
2x – 5 = 11
3x2 – 4x + 5 = 0
Expression: Combination of variables and constants which
evaluates to a single value
•
•
•
6x – 1
3x2 – x – 2
–8(x + 1)
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Your Turn
Determine whether the following are expressions or
equations.
a. 9x2 – 5x = 4
b. 3x + 2
c. 6(2x – 1) + 5
Solution:
a.9x2 – 5x = 4 is an equation
b.3x + 2 is an expression
c.6(2x – 1) + 5 is an expression
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2.
Determine whether a number is a
solution of an equation
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Solving an Equation
Given: x + 5 = 21
The letter x is called the variable (or the unknown).
Solution of the equation: A value for a variable which
makes the equation true
16 is a solution, because (16) + 5 = 21
3 is not a solution, because (3) + 5 ≠ 21
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Your Turn
Determine whether 6 is a solution of 3x – 5 = 2x.
Solution:
To see whether 6 is a solution, we can substitute 6 for x
and simplify.
3x – 5 = 2x
36–5≟ 26
18 – 5 ≟ 12
13 = 12
Substitute 6 for x.
Do the multiplication.
False.
Since 13 = 12 is a false statement, 6 is not a solution.
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3.
Solve a linear equation in one variable
by applying the addition or subtraction
property of equality
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Solving an Equation (Using add/subt prop)
Addition Property of Equality
Suppose that a, b, and c are real numbers. Then,
If a = b, then a + c = b + c.
Subtraction Property of Equality
Suppose that a, b, and c are real numbers. Then,
If a = b, then a – c = b – c.
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Solving an Equation
Comment
The subtraction property of equality is a special case of the
addition property. Instead of subtracting a number from
both sides of an equation, we could add the opposite of the
number to both sides.
When we use the properties described above, the resulting
equation will have the same solution set as the original
one. We say that the equations are equivalent.
Equivalent Equations
Two equations are called equivalent equations when they
have the same solution set.
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Example
Solve: x – 5 = 2
Solution:
x–5=2
x–5+5=2+5
Add 5 to both sides of the equation.
x+0=7
x=7
Apply the additive inverse property.
Apply the additive identity property.
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Example
cont’d
We check by substituting 7 for x in the original equation
and simplifying.
x–5=2
7–5≟2
2=2
Substitute 7 for x.
True.
Since the previous statement is true, 7 is a solution. The
solution set of this equation is {7}.
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4.
Solve a linear equation in one variable
by applying the multiplication or division
property of equality
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Solving an Equation (Using mult/div property)
Multiplication Property of Equality
Suppose that a, b, and c are real numbers.
If a = b, then ca = cb.
Division Property of Equality
Suppose that a, b, and c are real numbers and c  0.
If a = b, then
=
.
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Example
Solve:
Solution:
Multiply both sides by 3.
3
= x and 3  12 = 36.
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5. Application 1
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Example – Buying Furniture
A sofa is on sale for $650. If it has been marked down
$325, find its regular price.
Solution:
We can let r represent the regular price and substitute 650
for the sale price and 325 for the markdown in the following
formula.
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Example 8 – Solution
cont’d
We can use the addition property of equality to solve the
equation.
650 = r – 325
650 + 325 = r – 325 + 325
975 = r
Add 325 to both sides.
650 + 325 = 975 and –325 + 325 = 0.
The regular price is $975.
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6. Application 2
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Application 2
A percent is the numerator of a fraction with a denominator
of 100.
For example,
percent (written as
) is the fraction
or the decimal 0.0625. In problems involving percent, the
word of usually means multiplication.
For example,
8,500.
of 8,500 is the product of 0.0625 and
of 8,500 = 0.0625  8,500
= 531.25
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Percent
In the statement
of 8,500 = 531.25, the percent
is
called a rate, 8,500 is called the base, and their product,
531.25, is called the amount.
Every percent problem is based on the equation
rate  base = amount.
Percent Formula
If r is the rate, b is the base, and a is the amount, then
rb = a
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Percent
Percent problems involve questions such as the following.
• What is 30% of 1,000?
We must find the amount.
• 45% of what number is 405?
We must find the base.
• What percent of 400 is 60?
We must find the rate.
When we substitute the values of the rate, base, and
amount into the percent formula, we will obtain an equation
that we can solve.
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Example 10
What is 30% of 1,000?
Solution:
In this example, the rate r is 30% and the base is 1,000.
We must find the amount.
We can substitute these values into the percent formula
and solve for a.
rb = a
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Example 10 – Solution
30%  1,000 = a
Substitute 30% for r and 1,000 for b.
0.30  1,000 = a
Change 30% to the decimal 0.30.
300 = a
cont’d
Multiply.
Thus, 30% of 1,000 is 300.
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7.
Solve an application involving percents
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Example – Investing
At a stockholders meeting, members representing 4.5
million shares voted in favor of a proposal for a mandatory
retirement age for the members of the board of directors. If
this represented 75% of the number of shares outstanding,
how many shares were outstanding?
Solution:
Let b represent the number of outstanding shares. Then
75% of b is 4.5 million.
We can substitute 75% for r and 4.5 million for a in the
percent formula and solve for b.
rb = a
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Example 13 – Solution
75%  b = 4,500,000
0.75b = 4,500,000
=
b = 6,000,000
cont’d
4.5 million = 4,500,000
Change 75% to a decimal.
To undo the multiplication of 0.75,
divide both sides by 0.75.
Divide.
There were 6 million shares outstanding.
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