Download Section 6.1

Chapter 6
College Algebra
!  Are
there any questions about
our previous section’s HW?
!  Are
there any course questions
before we begin today?
College Algebra
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Section 6.1
College Algebra
!  Solve
a system of two linear equations
in two variables by graphing
!  Solve
a system of two linear equations
in two variables using the substitution
method and the elimination method
!  Use
systems of two linear equations to
solve applied problems
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!  A
system of linear equations consists of
two or more linear equations.
" 1st equation
!  Looks like #
$2nd equation
!  A
solution of a system of linear
equations in two variables is an ordered
pair
€ that is a solution of both equations
in the system.
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3x – 2y = 2
x + 2y = 6
no
b) (2,2)
yes
a) (0, -1)
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!  Since
a solution of a system of
equations is a solution common to both
equations, it is also a point common to
the graphs of both equations.
!  To
find the solution of a system of two
linear equations: (steps)
1.  Graph each equation
2.  Identify the intersection
College Algebra
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Solve the system of
equations by graphing.
# 2x − y = 6
$
% x + 3y = 10
First, graph 2x – y = 6.
€Second, graph x + 3y = 10.
The lines APPEAR to
intersect at (4, 2).
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Although the solution to the system of equations
appears to be (4, 2), you still need to check the
answer by substituting x = 4 and y = 2 into the
two equations.
First equation:
2(4) – (2) = 8 – 2 = 6
True
Second equation:
(4) + 3(2) = 4 + 6 = 10
True
The point (4, 2) checks, so it is the solution of the
system.
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!  Neatly
drawn graphs can help
when guessing the solution, but
you must check to make sure.
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Solve the system of
equations by graphing.
# −x + 3y = 6
$
% 3x − 9y = 9
First, graph -x + 3y = 6.
(0, 2)
(-6, 0)
€Second, graph 3x – 9y = 9.
(0, -1)
(3, 0)
The lines are parallel.
No solution
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Solve the system of
equations by graphing.
# x = 3y −1
$
%2x − 6y = −2
First, graph x = 3y – 1.
(-1, 0)
(0, ⅓)
€ Second, graph 2x – 6y = -2.
The lines are coincident.
{(x,y) | 2x – 6y = -2}
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! 
There are three possible outcomes when
graphing two linear equations in a plane.
!  One point of intersection " one solution
! 
Parallel lines " no solution
! 
Coincident lines " infinitely many solutions
! 
If there is at least one solution, the system is
considered to be consistent.
! 
If the system defines distinct lines, the
equations are independent.
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Graph
Number of Solutions
Type of System
intersecting lines
one solution
Consistent
The equations are
independent.
parallel lines
no solutions
Inconsistent
The equations are
independent.
coincident lines
(same line)
infinitely many
solutions
Consistent
The equations are
dependent.
Two lines intersect at one point.
Parallel lines
Lines coincide
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! 
Steps for Substitution:
1.  Solve one of the equations for one variable (try
to solve for the variable with a coefficient of one)
2.  Substitute the expression from step 1 into the
other equation and solve this new equation
3.  Substitute the value from step 2 into one of your
original equations to complete the ordered pair
4.  Check the ordered pair solution to make sure it
satisfies the entire system
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Solve the system using the substitution method.
3x – y = 6
-4x + 2y = –8
The first equation can be easily solved for y.
-y = –3x + 6
y = 3x – 6
Subtract 3x from both sides
Multiply both sides by –1
Now, substitute this value for y in the second equation.
-4x + 2(3x – 6) = –8
-4x + 6x – 12 = –8
2x – 12 = –8
2x = 4
x=2
College Algebra
Replace y with result from first equation
Distribute
Simplify
Add 12 to both sides
Divide both sides by 2
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Substitute x = 2 into the first equation, which has already been
solved for y.
y = 3x – 6 = 3(2) – 6 = 6 – 6 = 0
The solution of the system is the ordered pair (2, 0).
Check the point (2, 0) in the original equations.
First equation:
3x – y = 6 3(2) – 0 = 6
True
Second equation:
-4x + 2y = –8
-4(2) + 2(0) = –8
True
The solution of the system is (2, 0).
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Solve the system using the substitution method.
y = 2x – 5
8x – 4y = 20
The first equation is already solved for y.
Now, substitute this value for y in the second equation.
8x – 4(2x – 5) = 20
Replace y with result from first equation
8x – 8x + 20 = 20
Distribute
20 = 20
Simplify
This is a true statement, so these lines are the same.
Hence, the system has infinitely many solutions.
College Algebra
{(x,y) | 8x – 4y = 20}
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Solve the system using the substitution method.
3x – y = 4
6x – 2y = 4
The first equation can be easily solved for y.
-y = -3x + 4
Subtract 3x from both sides
y = 3x – 4
ultiply both sides by –1
M
Now, substitute this value for y in the second equation.
6x – 2(3x – 4) = 4
Replace y with result from first equation
6x – 6x + 8 = 4 Distribute
8=4 Simplify
This is a false statement, so these lines are the parallel.
Hence, the system has no solutions.
College Algebra
No solution
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! 
Steps for elimination:
1.  Make sure system s equations are written in
“pseudo-standard” form
2.  Make one of the variables have opposite
coefficients (multiply by a constant if necessary)
3.  Add the equations together and solve for the
remaining variable
4.  Substitute the value from step 3 into one of the
original equations to complete the ordered pair
5.  Check the ordered pair solution to make sure it
satisfies the entire system
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Solve the following
system by elimination 3x = y + 4
6x – 2y = 4
1
3x – y = y – y + 4
6x – 2y = 4
2
-2(3x – y = 4)
(6x – 2y = 4)
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-6x + 2y = -8
6x – 2y = 4
0 + 0 = -4
0 =
≠ -4
False!
No Solution
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Solve the following
system by elimination 3
8x + 3y = -18
⅔x + ¼y = -3/2
6x – 3y = -24
½x – ¼y = -2
14x + 0 = -42
14x = -42
2
1
14
14
x = -3
12(⅔x + ¼y = -3/2)
12(½x – ¼y = -2)
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4
Use x = -3 to find y
1st equation:
5
8x + 3y = -18 ⅔(-3) + ¼(2) = -3/2
8(-3) + 3y = -18
-2 + ½ = -3/2
True
-24 + 3y = -18
½(-3) – ¼(2) = -2
+24
+24
-3/2 – ½ = -2
3y = 6
True
3
3
y = 2
So our solution is
College Algebra
(-3, 2) ?
(-3, 2)
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Solve the following
system by elimination 3
-8x + 4y = -20
8x
–
4y
=
20
y = 2x – 5
0 + 0 = 0
8x – 4y = 20
0 =
= 0
-2x + y = 2x – 2x – 5
True!
1
8x – 4y = 20
2
4(-2x + y = -5)
(8x – 4y = 20)
College Algebra
{(x,y) | 8x – 4y = 20}
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General Strategy for Problem Solving
1. U
NDERSTAND the problem.
•  Read and reread the problem
•  Choose two variables to represent the two
unknowns
•  Construct a drawing, whenever possible
2. MODEL the problem with a system.
3. SOLVE the system of equations.
4. INTERPRET the result.
•  Check proposed solution in original
problem.
•  State your conclusion.
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Hilton University Drama club sold 311 tickets
for a play. Student tickets cost 50 cents
each; non-student tickets cost $1.50. If the
total receipts were $385.50, find how many
tickets of each type were sold.
1. UNDERSTAND
Since we are looking for two numbers, we let
s = the number of student tickets
n = the number of non-student tickets
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2. TRANSLATE
Hilton University Drama club sold 311 tickets for a play.
s + n = 311
total receipts were $385.50
Admission for
students
0.50s
College Algebra
Admission for
non-students
+
1.50n
Total
receipts
=
385.50
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3. SOLVE
We are solving the system s + n = 311
0.50s + 1.50n = 385.50
We should use the elimination method since the
equations are in pseudo-standard form, and we need to
multiply to clear the decimals anyway.
s + n = 311
–2(0.50s + 1.50n) = –2(385.50)
College Algebra
s + n = 311
–s – 3n = –771
–2n = –460
n = 230
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Now we substitute 230 for n into the first equation to
solve for s.
s + n = 311
s + 230 = 311
s = 81
4. I NTERPRET
Check: Substitute s = 81 and n = 230 into both of the
equations.
s + n = 311
First Equation
81 + 230 = 311
0.50s + 1.50n = 385.50
0.50(81) + 1.50(230) = 385.50
40.50 + 345 = 385.50 True
Second Equation
True
State: There were 81 student tickets and 230 non
student tickets sold.
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Terry Watkins can row about 10.6 kilometers in 1 hour
downstream and 6.8 kilometers upstream in 1 hour. Find
how fast he can row in still water, and find the speed of
the current.
1. UNDERSTAND
Remember that d = r • t (or r • t = d)
But what about the water?
Since we are looking for two rates, we let
r = the rate of the rower in still water
w = the rate of the water current
So, when traveling downstream, add the speeds " r + w when traveling upstream, subtract the water s speed " r – w
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2. TRANSLATE
rate
downstream
(r + w) time
downstream
•
rate
upstream
(r – w) College Algebra
1
distance
downstream
=
time
upstream
•
1
10.6
distance
upstream
=
6.8
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3. SOLVE
We are solving the system r + w = 10.6
r – w = 6.8
We should use the elimination method since the two
equations are in standard form.
r + w = 10.6
r – w = 6.8
2r = 17.4
r = 8.7
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Now we substitute 8.7 for r into the first equation.
4. INTERPRET
r + w = 10.6
8.7 + w = 10.6
w = 1.9
Check: Substitute r = 8.7 and w = 1.9 into both equations.
(r + w)1 = 10.6
(8.7 + 1.9)1 = 10.6
(r – w)1 = 1.9
First equation
True
Second equation
(8.7 – 1.9)1 = 6.8
True
State: Terry s rate in still water is 8.7 km/hr
and the rate of the water current is 1.9
km/hr.
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A Candy Barrel shop manager mixes M&M s worth $2.00
per pound with trail mix worth $1.50 per pound. How
many pounds of each should she use to get 50 pounds of a
party mix worth $1.80 per pound?
1. UNDERSTAND
To find out the cost of any quantity of items we use
the formula
price per unit
•
number of units
=
price of all units
Since we are looking for two quantities, we let
x = the amount of M&M s
y = the amount of trail mix
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2. TRANSLATE
Fifty pounds of party mix
x + y = 50
price per unit
•
Price of
M&M s
2x
College Algebra
number of units
=
Price of
mixture
Price of
trail mix
+
1.5y
price of all units
=
1.8(50)
90
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3. SOLVE
We are solving the system x + y = 50
2x + 1.50y = 90
We should use the elimination method since the
equations are in pseudo-standard form, and we need to
multiply to clear the decimals anyway.
3(x + y) = 3(50)
–2(2x + 1.50y) = –2(90)
College Algebra
3x + 3y = 150
–4x – 3y = –180
–x = –30
x = 30
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Now we substitute 30 for x into the first equation.
x + y = 50
30 + y = 50
y = 20
4. INTERPRET
Check: Substitute x = 30 and y = 20 into both of the
equations.
x + y = 50 First equation
30 + 20 = 50 T
rue
2x + 1.50y = 90 S
econd equation
2(30) + 1.50(20) = 90
60 + 30 = 90 T
rue
State: The store manager needs to mix 30
pounds of M&M s and 20 pounds of trail
mix to get the mixture at $1.80 a pound.
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When should the substitution
method be used?
a coefficient is one or an equation is
already solved for a variable
When should the elimination method
be used?
College Algebra
coefficients are not one
and/or opposite signs
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