Download January 8 (§4.1): Systems of Two Equations in Two Variables

Systems of Two Equations in Two Variables
MATH 107: Finite Mathematics
University of Louisville
January 6 8, 2014
Introduction
Denitions
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What is a system of linear equations?
A linear equation in two variables is an equation of the form
ax + by = c , with numbers replacing a, b, and c . Our variables are x
and y .
A system of two linear equations is a pair of equations linear
equations, conventionally paired with a curly-brace, as such:
(
ax + by = c
dx + ey = f
with numbers replacing a, b, c , d , e , and f .
A solution of a system of linear equations is a pair (x , y ) of values
which makes both equations true.
MATH 107 (UofL)
Notes
January 8, 2014
Introduction
Denitions
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Example and Verication
Here is a system of equations:
(
4 11
2
4
x− y=
x+ y=
Is (15, 1) a solution of this system? No! Even though 15 − 4 · 1 = 11,
2 · 15 + 1 6= 2.
Is (3, −2) a solution of this system? Yes! Both 3 − 4(−2) = 11 and
2 · 3 + (−2) = 4.
But how could we know to check (3, −2) in the rst place?
MATH 107 (UofL)
Notes
Introduction
January 8, 2014
Motivation
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Applications
Several questions are solvable with the use of systems of linear
equations:
I have 18 coins, each of which is a penny or a nickel. My
collection is worth 42 cents. How many of each coin do I have?
A museum wants to admit two adults and a child for $20, and
one adult and three children for $22.50. What admission rates
should they charge?
You want to buy a kilogram of bulk candy for $8. Caramels weigh
10 grams and cost 5 cents apiece. Chocolates weigh 5 grams and
cost 7 cents apiece. How many of each candy should you buy?
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MATH 107 (UofL)
Notes
January 8, 2014
Introduction
Motivation
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Application #1: Coins
Question
I have 18 coins, each of which is a penny or a nickel. My collection is
worth 42 cents. How many of each coin do I have?
Two unknown quantities: number of pennies, and number of nickels.
Let's denote them x and y .
So we have x + y coins in total, with a value of 1x + 5y cents.Thus
we'll want a solution to the system of equations:
(
x + y = 18 ←− number of coins
x + 5y = 42 ←− value of coins
MATH 107 (UofL)
Notes
Introduction
January 8, 2014
Motivation
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Application #2: Museum admissions
Question
A museum wants to admit two adults and a child for $20, and one
adult and three children for $22.50. What admission rates should they
charge?
Two unknown quantities: cost of adult admission in dollars, and cost
of child admission in dollars.
Let's denote them x and y .
So two adults and a child cost 2x + y dollars, and one adult and three
children cost x + 3y dollars.Thus we'll want a solution to the system
of equations:
(
2x + y = 20 ←− cost of 2 adults and a child
x + 3y = 22.50 ←− cost of an adult and 3 children
MATH 107 (UofL)
Notes
January 8, 2014
Introduction
Motivation
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Application #3: Candy purchases
Question
You want to buy 1000 grams of bulk candy for 800 cents. Caramels
weigh 10 grams and cost 5 cents apiece. Chocolates weigh 5 grams
and cost 7 cents apiece. How many of each candy should you buy?
Two unknown quantities: quantity of caramels purchased, and
quantity of chocolates purchased.
Let's denote them x and y .
So our purchase will have a weight of 10x + 5y grams, and a price of
5x + 7y cents.Thus we'll want a solution to the system of equations:
(
10x + 5y = 1000 ←− weight of the candy in grams
5x + 7y = 800 ←− cost of the candy in cents
MATH 107 (UofL)
Notes
January 8, 2014
Solving Systems of Equations
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How to Solve!
Once we have systems of equations, we want to nd their solutions.
There are several approaches to doing so.
Graphing
Algebraic substitution
Algebraic elimination
Matrix methods
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MATH 107 (UofL)
Notes
January 8, 2014
Solving Systems of Equations
Graphing
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Graphing a Linear Equation
Recall that a linear equation is so called because its graph is a line.
Linear equations can have graphs like any of the three forms below:
c
b
b
a
c
a
c
a
ax + by = c
by = c
ax = c
A system of two equations, then, can be represented by a pair of lines.
MATH 107 (UofL)
Notes
Solving Systems of Equations
January 8, 2014
Graphing
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Graphing and Interpreting a System
4 11
2
4
The rst line passes through (11, 0) and (0, ).
The second line passes through (2, 0) and (0, 4).
And they intersect at the point (3, −2), so that's the solution!
(
x− y=
x+ y=
−11
4
MATH 107 (UofL)
Notes
January 8, 2014
Solving Systems of Equations
Graphing
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Graphing in Context
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Advantages:
•
•
•
Highly visual
Easily interpreted
Can be done with almost no algebraic skill
•
•
•
Requires tools
May be dicult to do exactly
Non-integer solutions are dicult to interpret
Disadvantages:
MATH 107 (UofL)
Solving Systems of Equations
Notes
January 8, 2014
Substitution
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Single-Variable Solution
A linear equation in a single variable is easy to solve, for example:
5x + 3 = 19
−3 −3
5x = 11
÷5
= ÷5
11 = 2.2
x
=
5
We can use the same technique to solve for the variables in a system
of equations one at a time.
MATH 107 (UofL)
Notes
January 8, 2014
Solving Systems of Equations
Substitution
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The Substitution Method
Let's consider our old friend:(
4 11
2
4
We can take one of these equations and solve for one variable; let's
take the second equation and solve for x .
x− y=
x+ y=
2x + y = 4
− 2x
− 2x
y = 4 − 2x
Now we can substitute this relationship back into the rst equation:
x − 4y = 11
x − 4(4 − 2x ) = 11
MATH 107 (UofL)
Solving Systems of Equations
Notes
January 8, 2014
Substitution
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The Substitution Method, continued
The equation at the end of the last slide is a linear equation in one
variable.
x − 4(4 − 2x ) = 11
x − 16 + 8x = 11
9x − 16 = 11
9x = 27
x =3
And now,from the rule that y = 4 − 2x , we can nd that
y = 4 − 2 · 3 = −2.
MATH 107 (UofL)
Notes
January 8, 2014
Solving Systems of Equations
Substitution
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The General Approach
To perform substitution, go throguh the following steps:
1. Choose one of the two equations, and solve it for one of the
variables.
2. Substitute the variable you solved for into the second equation.
3. Solve the second equation, which is now in terms of a single
variable.
4. Substitute the solution you nd back into your solution from step
1.
In step one you seem to be making an arbitrary decision. But
whichever choice you make will work.
MATH 107 (UofL)
Notes
Solving Systems of Equations
January 8, 2014
Substitution
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The Same Story, Told Twice
(
4 11
2
4
x− y=
x+ y=
x − 4y = 11
2x + y = 4
y=
y = 4 − 2x
2x + y = 4
x − 4y = 11
2x + = 4
x − 4(4 − 2x ) = 11
2x + x − = 4
x − 16 + 8x = 11
9x − 16 = 11
x− =4
9x = 27
x=
x =3
x =3
y = 4 − 2x
y=
y = 4 − 2 · 3 = −2
y=
x −11
4
1
4
9
4
x −11
4
11
4
11
4
9
4
27
4
x −11
4
3−11
4
MATH 107 (UofL)
Notes
2
=−
January 8, 2014
Solving Systems of Equations
Elimination
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Equation-Manipulation Techniques
In a system of two equations, the following operations don't change
the solution of the system:
Swapping the two equations
Multiplying one equation by a nonzero number
Adding one equation to the other
In addition, using the second and third operation, we can also:
Add a multiple of one equation to another
We can eliminate a variable from one equation with these operations.
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MATH 107 (UofL)
Solving Systems of Equations
Notes
January 8, 2014
Elimination
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An elimination example
Let's return to our previously solved problem:
(
x − 4y = 11
2x + y = 4
We can multiply the rst equation by −2 to get two equations which,
when added together, will have no x :
(
−2x + 8y = −22
2x + y = 4
And now we can add the second equation to the rst:
(
9y = −18
2x + y = 4
MATH 107 (UofL)
Notes
January 8, 2014
Solving Systems of Equations
Elimination
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An elimination example, continued
9y = −18
2x + y = 4
Now, we can multiply the rst
equation
by
:
(
y = −2
2x + y = 4
And subtract it from the second
equation:
(
y = −2
2x = 6
Finally, we divide the second(equation by 2:
y = −2
x
= 3
(
1
9
MATH 107 (UofL)
Notes
January 8, 2014
Unsolvable Systems
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Contradictory systems
Sometimes, these methods (don't produce a solution:
2x − 4y = 8
−x + 2y = 5
2x − 4y = 8
2 5
−x + 2y = 5
y=
(
2x − 4y = 8
x − 2y = 4
2x − 4 = 8
−x + 2y = 5
(
2x − 2x − 10 = 8
x − 2y = 4
−10 = 8??
0 = 9??
With both approaches, we arrive at a contradiction, so this equation
has no solution.
(
−x + y =
x +5
2
x +5
2
MATH 107 (UofL)
Notes
January 8, 2014
Unsolvable Systems
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Underspecied systems
Sometimes there's no contradiction,
but not a unique solution either:
(
2x − 4y = 8
−x + 2y = −4
2x − 4y = 8
2 4
−x + 2y = −4
y=
(
2x − 4y = 8
x − 2y = 4
2x − 4 = 8
−x + 2y = −4
(
2x − 2x + 8 = 8
x − 2y = 4
8 = 8??
0 = 0??
With both approaches, we arrive at a tautology, so this equation has
innitely many solutions.
(
−x + y = −
x −4
2
x −4
2
MATH 107 (UofL)
Notes
January 8, 2014
Summary and Examples
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Summary of Techniques and Interpretation
We have seen three approaches to interpreting and solving systems:
Graphing not recommended for nding solutions!
Substitution
Elimination
One of three results can arise:
Completed process or two intersecting lines (one solution)
Contradictory result or two parallel lines (no solution)
Tautological result or two coincident lines (many solutions)
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MATH 107 (UofL)
Notes
January 8, 2014
Summary and Examples
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Application #1 revisited
(
18 ←− number of coins
5 42 ←− value of coins
x+ y=
x+ y=
(
18
x + y = 18
18
x + 5y = 42
(
5 42
x + y = −18
5 18
42
4
y = 24
(
90 5 42
−x − y = −18
4 48
y =6
12
(
18 12 −x = −12
y =6
6
Thus, we have 12 pennies and 6 nickels.
x +y =
y = −x
x+ y=
x + ( − x) =
x+ − x=
− x =−
x=
y= −
y=
MATH 107 (UofL)
Notes
12 6
( , )
January 8, 2014
Summary and Examples
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Application #2 revisited
20 ←− cost of 2 adults and a child
3 22.50 ←− cost of an adult and 3 children
(
2x + y = 20
2x + y = 20
y = 20 − 2x
2x + 6y = 45
(
x + 3y = 22.5
(7.50, 5)
2
x + y = 20
x + 3(20 − 2x ) = 22.5
5y = 25
(
x + 60 − 6x = 22.5
2x + y = 20
−5x = −37.5
y =5
x = 7.50
(
2x = 15
y = 20 − 2 · 7.5
y =5
y =5
Thus, the price will be $7.50 for an adult and $5.00 for a child.
(
2x +
y=
x+ y=
MATH 107 (UofL)
Notes
January 8, 2014
Summary and Examples
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Application #3 revisited
10x + 5y = 1000 ←− weight of the candy in grams
5x + 7y = 800 ←− cost of the candy in cents
(
10x + 5y = 1000
10x + 5y = 1000
10x + 14y = 1600
y = 200 − 2x (
10x + 5y = 1000
5x + 7y = 800
9y = 600
5x + 7(200 − 2x ) = 800
(
5x + 1400 − 14x = 800
10x + 5y = 1000
−9x = 600
y=
(
x=
≈ 67
10x =
y=
≈ 67
y=
Thus, we want about 67 chocolates and 67 caramels.
(
200
3
2000
3
200
3
200
3
200
3
MATH 107 (UofL)
Notes
January 8, 2014