Chapter 1
Linear Equations
and Graphs
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 1
Linear Equations, Standard Form
In general, a first-degree, or linear, equation in one variable
is any equation that can be written in the form
Standard Form :
ax  b  0, a  0
The equation
x
3  2( x  3)   5
3
is a first degree equation in one variable.
This equation can be converted to standard form
by clearing fractions and simplifying.
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 2
2
Theorem 1: Equality Properties
An equivalent equation will result if
1. The same quantity is added to or subtracted from each
side of a given equation.
2. Each side of a given equation is multiplied by or divided
by the same nonzero quantity.
To solve a linear equation, we perform these operations on
the equation to obtain simpler equivalent forms, until we
obtain an equation with an obvious solution.
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
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Slide 3
3
Example 1: Solve the Linear Equation
and Check the Result
x2 x
Example : Solve
 5
2
3
 x2 x
Solution : 6 
   65
3
 2
3( x  2)  2 x  30
3 x  6  2 x  30
x  6  30
x  24
Check: When x  24,
x  2 x 24  2 24 26 24
- 


 13  8  5
2
3
2
3
2
3
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
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Slide 4
4
Example 2: Solving a Formula for
a Particular Variable
Solve the equation M =Nt +Nr for N.
Solution:
M  N (t  r )
Factor N .
M
N
tr
Divide by (t  r ).
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 5
Linear Inequalities
For real numbers, a and b, a < b (a is less than b)
if there exists a positive number p such that a + p = b.
a < b is equivalent to b > a
(b is greater than a).
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 6
Example 3: Inequalities
Replace each question mark with either < or > and find a
value p that, when added to the lesser value results in an
equation.
(A) 4 ? 12
Solution:
(A) 4 < 12
(B) –10 ? –3
(C) 0 ? –19
(B) –10 < –3
p = 7 gives –10 + 7 = –3.
(C) 0 > –19
p = 19 gives – 19 + 19 = 0.
p = 8 gives 4 + 8 = 12.
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 7
Theorem 2: Inequality Properties
An equivalent inequality will result, and the sense or direction
will remain the same if each side of the original inequality
1. has the same real number added to or subtracted from it.
2. is multiplied or divided by the same positive number.
An equivalent inequality will result, and the sense or direction
will reverse if each side of the original inequality
3. is multiplied or divided by the same negative number.
Note: Multiplication by 0 and division by 0 are not permitted.
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 8
Interval Notation
Interval notation gives a convenient way to represent
collections of numbers that are given as inequalities or as
number line graphs.
The table on the following slide summarizes comparisons of
the various representations of collections of numbers in
interval notation, inequality notation, and line graphs.
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 9
Interval Notation Table
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Slide 10
Interval Notation Continued
In the table, the numbers a and b, with a < b, are called the
endpoints of the interval.
An interval is closed if the interval contains its endpoints.
A square bracket is used to indicate that an endpoint is
included in the interval.
An interval is open if the interval does not contain any of its
endpoints.
Parentheses are used to indicate that an endpoint is not
included in the interval.
Parentheses are always used for negative or positive
infinity.
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 11
Example 4: Interval and Inequality
Notation and Line Graphs
(A) Write [–5, 2) as a double inequality and graph .
(B) Write x ≥ –2 in interval notation and graph.
Solution:
[–5, 2) is equivalent to –5 ≤ x < 2
[
)
–5
2
(B) x ≥ –2 is equivalent to –2 < x which is the interval [–2, ∞).
[
–2
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Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 12
First-Degree Linear Inequality
If, in a linear equation, the equality symbol is replaced by
<, >, <, or >, the resulting expression is called a firstdegree, or linear, inequality.
The inequality
x
 2(3 x  1)  5
2
is a first-degree (linear) inequality in one variable.
The inequality
x
5  1  3 x  2  is a linear inequality.
2
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Slide 13
13
Solving Linear Inequalities
We can perform the same operations on inequalities as equations,
except that the sense of the inequality reverses if we multiply
or divide both sides by a negative number.
Starting with the true statement –2 > –9, when we multiply both
sides by 3, we obtain –6 > –27.
The sense of the inequality remains the same.
If we multiply both sides by -3 instead, we must reverse the sense
of the inequality and write 6 < 27 to have a true statement.
The sense of the inequality reverses whenever we multiply or
divide across the inequality by a negative number.
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Slide 14
Example 5: Solve a
Linear Inequality
Solve the inequality 3(x – 1) < 5(x + 2) – 5.
Solution:
3(x –1) < 5(x + 2) – 5
3x – 3 < 5x + 10 – 5 Distribute the 3 and the 5.
3x – 3 < 5x + 5
Combine like terms.
–2x < 8
Subtract 5x and add 3 to both sides.
x > –4
Divide both sides by –2 and reverse
the inequality.
Interval notation for this solution is (–4, ∞)
The number line graph is
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Slide 15
Example 6: Solve a
Double Linear Inequality
Solve and give the solution in interval notation –7 < 3x – 4 < 8
Solution: Isolate the variable x in the middle.
–7 < 3x – 4 < 8
– 3 < 3x < 12
Add four to all parts.
3 3 x 12


3
3
3
Divide by 3.
–1 < x < 4
Simplify.
The solution in interval form is [–1, 4).
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 16
Profit–loss Analysis
 Companies use Profit–loss analysis to support decisions
regarding the pricing of products and appropriate levels of
production in order to maximize company profit.
 A manufacturing company has costs, C, which include fixed
costs (plant overhead, product design, setup, and promotion)
and variable costs (costs that depend on the number of
items produced.)
 The revenue (income) for a company, R, is the amount of
money the company receives from selling its product.
• If R < C, the company loses money.
• If R = C, the company breaks even.
• If R > C, the company makes a profit.
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 17
Profit–loss Analysis
 Profit, P, is equal to revenue, R, minus cost, C.
• P = R – C.
 When P < 0, the company loses money (cost exceeds revenue).
 When P = 0, the company breaks even (cost equals revenue).
 When P > 0, the company makes a profit (revenue exceeds
cost).
 Often it is helpful to find when the company breaks even. This
is called break-even analysis.
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 18
Example 7: Break-Even Analysis
A university club plans to raise money by selling custom
printed t-shirts. They find that a printer charges $400 for
creating the art work and $3.50 per shirt that is printed.
If they sell the shirts for $15.00 each, how many shirts must
they make and sell to break even.
Solution: Let n represent the number of shirts made and sold.
The cost for having n shirts printed is C(n) = 400 + 3.50n.
When n shirts are sold, the revenue is R(n) = 15n.
Break-even is when cost equals revenue.
Solve the equation: 400 + 3.50n = 15n.
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 19
Example 7: Break-Even Analysis
(continued)
Solve the equation: 400 + 3.50n = 15n
400 + 3.50n – 3.50n= 15n – 3.50n Subtract 3.50n.
400 = 11.50n
Simplify.
400
n
11.50
Divide by 11.50.
n = 34.78 ≈ 35.
Round to a whole number value.
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 20
Example 7: Break-Even Analysis
(continued)
Check: If 34 shirts are made, cost is 400 + 3.5 (34) = $519.00
If 34 shirts are sold for $15 each, revenue is $510.00.
When 34 shirts are made and sold, cost exceeds revenue.
If 35 shirts are made, the cost is 400 + 3.5 (35) = $522.50
If 35 shirts are sold for $15 each, the revenue is $525.00.
When 35 shirts are made and sold, the revenue exceeds the cost so
there is a profit. The break even point is between 34 and 35
shirts. It does not make practical sense to consider a portion of
a shirt so the break-even point is rounded to whole number
values.
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 21
Section 1.2 Graphs and Lines
In this section, we examine one of the most basic geometric
figures—a line.
When the term line is used, we mean straight line.
This section will outline how to recognize and graph a line
and how to use information concerning a line to find its
equation.
Examining graphs of equations often reveals additional
insights into the nature of the equation’s solutions.
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 22
The Cartesian Coordinate System
The Cartesian or rectangular coordinate system consists of
two real number lines, one horizontal and one vertical, that
intersect at their origins.
The two number lines are called the horizontal axis and the
vertical axis.
Together, they are referred to as the coordinate axes.
The horizontal axis is usually referred to as the x axis and the
vertical axis as the y axis.
The coordinate axes divide the plane into four parts called
quadrants, which are numbered counterclockwise using
Roman numerals from I to IV, starting with the upper right
quadrant.
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 23
The Cartesian Coordinate System
Labeled components of the Cartesian coordinate system are
shown in the figure.
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 24
The Fundamental Theorem
of Analytic Geometry
There is a one-to-one correspondence between the points in a
plane and the elements in the set of all ordered pairs of real
numbers.
Each point in the Cartesian coordinate system corresponds to
exactly one ordered pair of real numbers.
Each ordered pair of real numbers corresponds to exactly one
point in the Cartesian coordinate system.
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 25
Graphs of Ax + By = C
Earlier, we called an equation of the form ax + b = 0 (a ≠ 0) a
linear equation in one variable.
We now consider linear equations in two variables:
DEFINITION Linear Equations in Two
Variables
A linear equation in two variables is an equation that can
be written in the standard form
Ax + By = C
where A, B, and C are constants (A and B not both 0), and x
and y are variables.
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 26
Solutions to Equations
in Two Variables
Consider the standard form linear equation 2x – 3y = 6.
A solution to an equation in two variables is an ordered pair
of numbers that when substituted into the equation result in a
true statement.
The ordered pairs (x, y) = (3, 0) and (x, y) = (9, 4) are
solutions to the equation. When values in these ordered pairs
are substituted into the equation, the result is a true statement.
The ordered pair (x, y) = (0, 2) is not a solution to the
equation. When values in this ordered pair are substituted into
the equation, the resulting statement is not true.
However, the ordered pair (x, y) = (0, –2) is a solution.
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 27
Theorem 1: Graph of a Linear
Equation in Two Variables
The graph of any equation of the form
Ax + By = C (A and B not both 0)
is a line, and any line in a Cartesian coordinate system is the
graph of an equation of this form.
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 28
Theorem 1 Observations
If A  0 and B  0, then the standard form linear equation
Ax  By  C can be written as
A
C
y   x   mx  b, m  0
B
B
If A  0 and B  0, then the standard form linear equation
C
can be written as y 
and its graph is a horizontal line.
B
If A  0 and B  0, then the standard form linear equation
C
can be written as x 
and its graph is a vertical line.
A
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Slide 29
Graphing a Linear Equation in
Two Variables
To graph a linear function in two variables, plot any two
points in the solution set and use a straightedge to draw the
line through these two points.
The points where the line crosses the axes are often the
easiest points to find since these allow substitution of x = 0
and/or y = 0.
The point corresponding to x = 0 is a point on the y axis and is
called the y intercept.
The point corresponding to y = 0 is a point on the x axis and is
called the x intercept.
Note: It is a good idea to plot a third solution point to check
your work.
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Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 30
Example: Using Intercepts to
Graph a Line
Graph: 3x – 4y = 12
Solution: Find and plot the y intercept, the x intercept, and a
check point.
The y intercept is found by letting x = 0 and solving for y.
This gives 3·0 – 4y = 12 which gives y = –3.
The y intercept is the point (0, –3).
The x intercept is found by letting y = 0 and solving for x.
This gives 3x – 4·0 = 12 which gives x = 4.
The x intercept is the point (4, 0).
A third point can be found by letting x = 8 and solving to find
y = 3 for the point (8, 3).
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Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 31
Using Intercepts to Graph a Line
continued
The three points are plotted on the graph and a straight line is
drawn through the three points as shown in the figure.
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Slide 32
Example: Graph a Line Using a
Graphing Calculator
Graph: 3x – 4y = 12 on a graphing calculator and find the
intercepts.
Solution: Solve 3x – 4y = 12 for y.
3x – 4y = 12
Add –3x to both sides.
– 4y = –3x + 12 Divide both sides by –4.
3 x  12
Simplify.
y
4
3
y  x3
4
The right side of this equation is entered into the calculator.
https://www.geogebra.org/calculator
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 33
Example: Horizontal and
Vertical Lines
Graph x = –4 and y = 6 simultaneously in the same rectangular
coordinate system.
Solution: the line x = –4
consists of all points with x
coordinate –4. This graph is the
vertical line through (–4, 0).
The line y = 6 consists of all
points with y coordinate 6. To
graph it, draw the horizontal
line through (0, 6).
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 34
Example: Horizontal
and Vertical Lines
Write the equations of the vertical and horizontal lines that
pass through the point (7, –5).
Solution:
The horizontal line through (7, –5) has equation, y = –5.
The vertical line through (7, -5) has equation x = 7.
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 35
The Slope of a Line
For two points, (x1, y1) and (x2, y2), on a line, the ratio of the
change in y to the change in x is called the slope of the line.
In a sense, slope measures the “steepness” or “tilt” of a line
relative to the x axis.
The change in x is often called the run.
The change in y is often called the rise.
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Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 36
Definition: Slope of a Line
If a line passes through two distinct points, (x1, y1) and
(x2, y2), then its slope is given by the formula
y2  y1
m
for x1  x2
x2  x1
vertical change (rise)

horizontal change (run)
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Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 37
Geometric Interpretation of
Positive Slope
When a line rises as x moves from left to right, the slope is a
positive value.
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 38
Geometric Interpretation of
Negative Slope
When a line falls as x moves from left to right, the slope is a
negative value.
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 39
Geometric Interpretation of Slope
Equal to Zero
A line that neither rises nor falls as x moves from left to right
has a slope equal to zero and is a horizontal line.
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 40
Geometric Interpretation of a
Line With Undefined Slope
A vertical line does not change horizontally but rises
indefinitely and is a line with undefined slope.
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 41
Example: Finding Slopes
Sketch a line through the pair of points, (–3, –2) and (3, 4)
and find the slope of the line.
Solution:
The graph rises as x increases.
The slope is positive.
The graph is shown.
4  ( 2) 6
m
 1
3  ( 3) 6
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Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 42
Example: Finding Slopes
Sketch a line through the pair of points, (–1, 3) and (2, –3)
and find the slope of the line.
Solution:
The graph falls as x increases.
The slope is negative.
The graph is shown.
3  3
6
m

 2
2  ( 1) 3
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Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 43
Example: Finding Slopes
Sketch a line through the pair of points, (–2, –3) and (3, –3)
and find the slope of the line.
Solution:
The graph neither falls nor
rises as x increases.
The graph is shown.
3  ( 3) 0
m
 0
3  ( 2) 5
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Slide 44
Example: Finding Slopes
Sketch a line through the pair of points, (–2, 4) and (–2, –2)
and find the slope of the line.
Solution:
The x value does not
change.
The graph is shown.
2  4
6
m

2  ( 2) 0
Slope is not defined.
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Slide 45
Equations of Lines: Special Forms
The linear equation form y = mx + b is called the slopeintercept form for a line.
When x = 0, then y = b and the graph of y = mx + b crosses
the y axis at the point (0, b). The constant b is the y intercept.
When x = 0, the point (0, b) is on the graph.
When x = 1, the point (1, m + b) is on the graph.
The slope of this line is given by:
y2  y1 ( m  b )  b
m

m
x2  x1
1 0
It follows that m is the slope of the line given by y = mx + b.
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Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 46
Definition: Slope-Intercept Form
The equation
y = mx + b (m = slope, b = y intercept)
is called the slope-intercept form of an equation of a line.
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Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 47
Example: Using the SlopeIntercept Form
Find the slope and y intercept, and graph y = –2x + 3.
Solution:
The slope of this line is m = –2; the y intercept is 3.
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Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 48
Example: Using the SlopeIntercept Form
Find the slope and y intercept, and graph y = –2x + 3.
Solution:
To graph this line, plot the y
intercept (0, 3).
Since the slope is –2, starting
at the y intercept, locate a
second point by moving 1
unit in the x direction (run)
and –2 units in the y direction
(rise).
Draw the line through these
two points.
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Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 49
Definition: Point-Slope Form
An equation of a line with slope m that passes through the
point (x1, y1) is
y – y1 = m(x – x1)
which is called the point-slope form of an equation of a line.
This form is useful since it allows finding an equation for a
line when the slope and coordinates of a point on the line are
known.
This form can also be used to find an equation for a line when
the coordinates of two points are known by using the two
points to find the slope and then using the point-slope form
with one of the known points.
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Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 50
Example 1: Using the
Point-Slope Form
Find an equation for the line with slope ½ that passes through
the point (2, -5).
Write the final answer in the form Ax + By = C.
Solution: Use y – y1 = m(x – x1) with slope m = ½ and the
point (x1, y1) = (2, -5).
y  ( 5)  1 2 ( x  2)
y  5  1 2 ( x  2) Multiply both sides by 2
2 y  10  x  2
 x  2 y  12 or x  2 y  12
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Slide 51
Example 2: Using the
Point-Slope Form
Find an equation for the line that contains the points (–3, 4)
and (4, –7).
Write the final answer in the form Ax + By = C.
Solution: To use y – y1 = m(x – x1) we first find the slope.
The slope is found using
y2  y1
7  4
11
m


x2  x1 4  ( 3)
7
Using one of the points, say (-3, 4) we have
11
y  4   ( x  ( 3)) Simplify and multiply both sides by 7.
7
7( y  4)  11( x  3) gives 7 y  28  11x  33
In standard form 11x  7 y  5
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Slide 52
Summary Table:
Equations of Lines
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Slide 53
Example: Application
A company finds that office equipment that was purchased for
$20,000 will have a scrap value of $2,000 after 10 years.
Using straight line depreciation (linear), find the linear
equation that relates value (V) in dollars to time (t) in years.
Solution: For t = 0, V = 20,000 and for t = 10, V = 2,000.
The depreciation line contains the ordered pairs (0, 20,000)
and (10, 2000).
We find the slope of the line using the slope formula.
y2  y1 2000  20000 18, 000
m


 1,800
x2  x1
10  0
10
Observe that the point (0, 20,000) gives the y (V(t)) intercept.
It follows that V(t) = –1,800t + 20,000.
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 54
Supply and Demand
In a free market, the price of a product is determined
by the relationship between supply and demand.
The price tends to stabilize at the point of
intersection of the demand and supply equations.
This point of intersection is called the equilibrium
point.
The corresponding price is called the equilibrium
price.
The common value of supply and demand is called
the equilibrium quantity.
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 55
Example: Supply and Demand
Use the barley market data in the following table to find:
(a) A linear supply equation of the form p = mx + b
(b) A linear demand equation of the form p = mx + b
(c) The equilibrium point.
Year
Supply
Mil bu
Demand
Mil bu
Price
$/bu
2002
340
270
2.22
2003
370
250
2.72
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 56
Supply and Demand
Example continued
Solution: (a) For a supply equation in the form p = mx + b,
we find two points of the form (x, p) on the supply line.
The table gives the points, (340, 2.22) and (370, 2.72) as
points on the supply line.
The slope of the line is
2.72  2.22 0.5
m

 0.0167
370  340 30
Using the point-slope form and the point (340, 2.22) the pricesupply equation is: p – p1 = m(x – x1)
p – 2.22 = 0.0167(x – 340)
p – 2.22 = 0.0167x – 5.678
p = 0.0167x – 3.458
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 57
Supply and Demand
Example continued
Solution: (b) For a demand equation in the form p = mx + b,
we find two points of the form (x, p) on the demand line.
The table gives the points (270, 2.22) and (250, 2.72) as points
on the demand equation.
The slope of the demand line is m 
2.72  2.22
.5

 0.025
250  270 20
Using the point-slope form and the point (270, 2.22) the pricedemand equation is: p – p1 = m(x – x1)
p – 2.22 = –0.025(x – 270)
p – 2.22 = –0.025x + 6.75
p = –0.025x + 8.97
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 58
Supply and Demand
Example continued
(c) We now graph the two equations using a graphing
calculator and use a calculator process to find the equilibrium
point.
Use the window settings shown and the intersection of graphs
solution process to obtain:
The equilibrium point is about (298, 1.52). This means that
the common value of supply and demand is 298 million
bushels at a price of $1.52 per bushel.
Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e
Copyright © 2019, 2015, 2011 Pearson Education, Inc.
Slide 59