Download Alg 3 PPT Notes 1.2 - Lancaster City Schools

1
Equations, Inequalities, and
Mathematical Modeling
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1.2
Linear Equations in One
Variable
Copyright © Cengage Learning. All rights reserved.
Objectives
 Identify different types of equations.
 Solve linear equations in one variable.
 Solve rational equations that lead to linear
equations.
 Find x- and y-intercepts of graphs of equations
algebraically.
 Use linear equations to model and solve
real-life problems.
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Equations and Solutions of
Equations
4
Equations and Solutions of Equations
An equation in x is a statement that two algebraic
expressions are equal.
For example
3x – 5 = 7,
x2 – x – 6 = 0,
and
are equations.
To solve an equation in x means to find all values of x for
which the equation is true. Such values are solutions.
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Equations and Solutions of Equations
For instance, x = 4 is a solution of the equation 3x – 5 = 7
because 3(4) – 5 = 7 is a true statement.
The solutions of an equation depend on the kinds of
numbers being considered.
For instance, in the set of rational numbers, x2 = 10 has no
solution because there is no rational number whose square
is 10. However, in the set of real numbers, the equation
has the two solutions
and
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Equations and Solutions of Equations
An equation that is true for every real number in the domain
of the variable is called an identity. For example
x2 – 9 = (x + 3)(x – 3)
Identity
is an identity because it is a true statement for any real
value of x.
The equation
Identity
where x  0, is an identity because it is true for any nonzero
real value of x.
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Equations and Solutions of Equations
An equation that is true for just some (but not all) of the real
numbers in the domain of the variable is called a
conditional equation.
For example, the equation
x2 – 9 = 0
Conditional equation
is conditional because x = 3 and x = –3 are the only values
in the domain that satisfy the equation.
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Equations and Solutions of Equations
The equation
2x + 4 = 6
is conditional because x = 1 is the only value in the domain
that satisfies the equation.
A contradiction is an equation that is false for every real
number in the domain of the variable. For example, the
equation
2x – 4 = 2x + 1
Contradiction
is a contradiction because there are no real values of x for
which the equation is true.
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Linear Equations in One Variable
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Linear Equations in One Variable
A linear equation has exactly one solution. To see this,
consider the following steps. (Remember that a  0)
ax + b = 0
ax = –b
Write original equation.
Subtract b from each side.
Divide each side by a.
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Linear Equations in One Variable
So, the equation
ax + b = 0
has exactly one solution,
To solve a conditional equation in x, isolate x on one side
of the equation by a sequence of equivalent equations,
each having the same solution(s) as the original equation.
The operations that yield equivalent equations come from
the properties of equality.
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Linear Equations in One Variable
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Example 1 – Solving a Linear Equation
a. 3x – 6 = 0
3x = 6
x=2
b. 5x + 4 = 3x – 8
2x + 4 = –8
2x = –12
x = –6
Original equation
Add 6 to each side.
Divide each side by 3.
Original equation
Subtract 3x from each side.
Subtract 4 from each side.
Divide each side by 2.
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Linear Equations in One Variable
After solving an equation, you should check each solution
in the original equation. For instance, you can check the
solution of Example 1(a) as follows.
3x – 6 = 0
3(2) – 6 ≟ 0
0=0
Write original equation.
Substitute 2 for x.
Solution checks.
Try checking the solution of Example 1(b).
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Rational Equations That Lead to
Linear Equations
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Rational Equations That Lead to Linear Equations
A rational equation is an equation that involves one or
more fractional expressions. To solve a rational equation,
find the least common denominator (LCD) of all terms and
multiply every term by the LCD.
This process will clear the original equation of fractions and
produce a simpler equation to work with.
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Example 3 – Solving a Rational Equation
Solve
Solution:
Original equation
Multiply each term by the LCD.
Simplify.
Combine like terms.
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Example 3 – Solution
cont’d
Divide each side by 13.
The solution is
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Rational Equations That Lead to Linear Equations
When multiplying or dividing an equation by a variable
expression, it is possible to introduce an extraneous
solution that does not satisfy the original equation.
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Finding Intercepts Algebraically
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Finding Intercepts Algebraically
You have learned to find x- and y- intercepts using a
graphical approach.
Because all the points on the x-axis have a y-coordinate
equal to zero, and all the points on the y-axis have an
x-coordinate equal to zero, you can use an algebraic
approach to find x- and y-intercepts, as follows.
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Example 5 – Finding Intercepts Algebraically
Find the x- and y-intercepts of the graph of each equation
algebraically.
a. y = 4x + 1
b. 3x + 2y = 6
Solution:
a. To find the x-intercept, set y equal to zero and solve for
x, as follows.
Write original equation.
y = 4x + 1
0 = 4x + 1
–1 = 4x
=x
Substitute 0 for y.
Subtract 1 from each side.
Divide each side by 4.
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Example 5 – Solution
So, the x-intercept is
cont’d
.
To find the y-intercept, set x equal to zero and solve for
y, as follows.
Write original equation.
y = 4x + 1
y = 4(0) + 1
Substitute 0 for x.
y=1
Simplify.
So, the y-intercept is (0, 1).
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Example 5 – Solution
cont’d
b. To find the x-intercept, set y equal to zero and solve for
x, as follows.
Write original equation.
3x + 2y = 6
3x + 2(0) = 6
3x = 6
Substitute 0 for y.
Simplify.
x=2
Divide each side by 3.
So, the x-intercept is (2, 0).
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Example 5 – Solution
cont’d
To find the y-intercept, set x equal to zero and solve for
y, as follows.
Write original equation.
3x + 2y = 6
3(0) + 2y = 6
2y = 6
y=3
Substitute 0 for x.
Simplify.
Divide each side by 2.
So, the y-intercept is (0, 3).
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Application
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Example 6 – Female Participants in Athletic Programs
The number y (in millions) of female participants in high
school athletic programs in the United States from 2000
through 2010 can be approximated by the linear model
y = 0.045t + 2.70,
0  t  10
where t = 0 represents 2000.
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Example 6 – Female Participants in Athletic Programs
cont’d
(a) Find algebraically and interpret the y-intercept of the
graph of the linear model shown in Figure 1.10.
(b) Use the linear model to
predict the year in which
there will be 3.42 million
female participants.
Female Participants in
High School Athletics
Figure 1.10
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Example 6(a) – Solution
To find the y-intercept, let t = 0 and solve for y, as follows.
y = 0.045t + 2.70
Write original equation.
= 0.045(0) + 2.70
Substitute 0 for t.
= 2.70
Simplify.
So, the y-intercept is (0, 2.70). This means that there were
about 2.70 million female participants in 2000.
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Example 6(b) – Solution
cont’d
Let y = 3.42 and solve for t, as follows
y = 0.045t + 2.70
Write original equation.
3.42 = 0.045t + 2.70
Substitute 3.42 for y.
0.72 = 0.045t
Subtract 2.70 from each side.
16 = t
Divide each side by 0.045.
Because t = 0 represents 2000, t = 16 must represent
2016. So, from this model, there will be 3.42 million female
participants in 2016.
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