Transcript linear equation in two unknowns

3
Systems of Linear
Equations and Matrices
Copyright © Cengage Learning. All rights reserved.
3.1
Systems of Two Equations in Two Unknowns
Copyright © Cengage Learning. All rights reserved.
Systems of Two Equations in Two Unknowns
Suppose you have $3 in your pocket to spend on snacks
and a drink.
If x represents the amount you’ll spend on snacks and y
represents the amount you’ll spend on a drink, you can say
that x + y = 3.
On the other hand, if for some reason you want to spend
$1 more on snacks than on your drink, you can also say
that x – y = 1. These are simple examples of linear
equations in two unknowns.
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Systems of Two Equations in Two Unknowns
Linear Equations in Two Unknowns
A linear equation in two unknowns is an equation that
can be written in the form
ax + by = c
with a, b, and c being real numbers. The number a is called
the coefficient of x and b is called the coefficient of y.
A solution of an equation consists of a pair of numbers: a
value for x and a value for y that satisfy the equation.
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Systems of Two Equations in Two Unknowns
Quick Example
In the linear equation 3x – y = 15, the coefficients are a = 3
and b = –1. The point (x, y) = (5, 0) is a solution, because
3(5) – (0) = 15.
In fact, a single linear equation such as 3x – y = 15 has
infinitely many solutions: We could solve for y = 3x – 15
and then, for every value of x we choose, we can get the
corresponding value of y, giving a solution (x, y).
These solutions are the points on a straight line, the graph
of the equation.
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Systems of Two Equations in Two Unknowns
In this section we are concerned with pairs (x, y) that are
solutions of two linear equations at the same time.
For example, (2, 1) is a solution of both of the equations
x + y = 3 and x – y = 1, because substituting x = 2 and
y = 1 into these equations gives 2 + 1 = 3 (true) and
2 – 1 = 1 (also true), respectively.
So, in the simple example we began with, you could spend
$2 on snacks and $1 on a drink.
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Systems of Two Equations in Two Unknowns
In the Example 1, you will see how to graphically and
algebraically solve a system of two linear equations in two
unknowns.
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Example 1 – Two Ways of Solving a System: Graphically and Algebraically
Find all solutions (x, y) of the following system of two
equations:
x+y=3
x – y = 1.
Solution:
We will see how to find the solution(s) in two ways:
graphically and algebraically. Remember that a solution is
a pair (x, y) that simultaneously satisfies both equations.
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Example 1 – Solution
cont’d
Method 1: Graphical
We already know that the solutions of a single linear
equation are the points on its graph, which is a straight line.
For a point to represent a solution of two linear equations, it
must lie simultaneously on both of the corresponding lines.
In other words, it must be a point where the two lines cross,
or intersect.
A look at Figure 1 should convince us
that the lines cross only at the point
(2, 1), so this is the only possible
solution.
Figure 1
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Example 1 – Solution
cont’d
Method 2: Algebraic
In the algebraic approach, we try to combine the equations
in such a way as to eliminate one variable.
In this case, notice that if we add the left-hand sides of the
equations, the terms with y are eliminated.
So, we add the first equation to the second (that is, add the
left-hand sides and add the right hand sides):
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Example 1 – Solution
cont’d
2x = 4
x = 2.
Now that we know that x has to be 2, we can substitute
back into either equation to find y. Choosing the first
equation (it doesn’t matter which we choose), we have
2+y=3
y=3−2
= 1.
We have found that the only possible solution is x = 2 and
y = 1, or (x, y) = (2, 1).
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Systems of Two Equations in Two Unknowns
Graphical Method for Solving a System of Two Equations
in Two Unknowns
Graph both equations on the same graph. (For example,
solve each for y to find the slope and y-intercept.) A point of
intersection gives the solution to the system.
To find the point, you may need to adjust the range of
x-values you use. To find the point accurately you may
need to use a smaller range (or zoom in if using
technology).
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Systems of Two Equations in Two Unknowns
Algebraic Method for Solving a System of Two Equations
in Two Unknowns
Multiply each equation by a nonzero number so that the
coefficients of x are the same in absolute value but opposite
in sign.
Add the two equations to eliminate x; this gives an equation
in y that we can solve to find its value.
Substitute this value of y into one of the original equations to
find the value of x. (Note that we could eliminate y first
instead of x if it’s more convenient.)
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Systems of Two Equations in Two Unknowns
We summarize the three possible outcomes we have
encountered.
Possible Outcomes for a System of Two Linear Equations
in Two Unknowns
1. A single (or unique) solution: This happens when the
lines corresponding to the two equations are distinct and
not parallel so that they intersect at a single point.
2. No solution: This happens when the two lines are
parallel. We say that the system is inconsistent.
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Systems of Two Equations in Two Unknowns
3. An infinite number of solutions: This occurs when the
two equations represent the same straight line, and we
say that such a system is redundant, or dependent. In
this case, we can represent the solutions by choosing
one variable arbitrarily and solving for the other.
In cases 1 and 3, we say that the system of equations is
consistent because it has at least one solution.
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Applications
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Example 5 – Blending
Acme Baby Foods mixes two strengths of apple juice. One
quart of Beginner’s juice is made from 30 fluid ounces of
water and 2 fluid ounces of apple juice concentrate. One
quart of Advanced juice is made from 20 fluid ounces of
water and 12 fluid ounces of concentrate. Every day Acme
has available 30,000 fluid ounces of water and 3,600 fluid
ounces of concentrate. If the company wants to use all the
water and concentrate, how many quarts of each type of
juice should it mix?
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Example 5 – Solution
In all applications we follow the same general strategy.
1. Identify and label the unknowns.
What are we asked to find? To answer this question, it is
common to respond by saying, “The unknowns are
Beginner’s juice and Advanced juice.” Quite frankly, this
is a baffling statement.
Just what is unknown about juice? We need to be more
precise:
The unknowns are (1) the number of quarts of
Beginner’s juice and (2) the number of quarts of
Advanced juice made each day.
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Example 5 – Solution
cont’d
So, we label the unknowns as follows: Let
x = number of quarts of Beginner’s juice made each day
y = number of quarts of Advanced juice made each day.
2. Use the information given to set up equations in the
unknowns.
This step is trickier, and the strategy varies from problem
to problem. Here, the amount of juice the company can
make is constrained by the fact that they have limited
amounts of water and concentrate.
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Example 5 – Solution
cont’d
This example shows a kind of application we will often see,
and it is helpful in these problems to use a table to record
the amounts of the resources used.
We can now set up an equation for each of the items listed
in the left column of the table.
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Example 5 – Solution
cont’d
Water: We read across the first row. If Acme mixes x quarts
of Beginner’s juice, each quart using 30 fluid ounces of
water, and y quarts of Advanced juice, each using 20 fluid
ounces of water, it will use a total of 30x + 20y fluid ounces
of water.
But we are told that the total has to be 30,000 fluid ounces.
Thus, 30x + 20y = 30,000. This is our first equation.
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Example 5 – Solution
cont’d
Concentrate: We read across the second row. If Acme
mixes x quarts of Beginner’s juice, each using 2 fluid
ounces of concentrate, and y quarts of Advanced juice,
each using 12 fluid ounces of concentrate, it will use a
total of 2x + 12y fluid ounces of concentrate.
But we are told that the total has to be 3,600 fluid ounces.
Thus, 2x + 12y = 3,600.
Now we have two equations:
30x + 20y = 30,000
2x + 12y = 3,600.
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Example 5 – Solution
cont’d
To make the numbers easier to work with, let’s divide
(both sides of) the first equation by 10 and the second by 2:
3x + 2y = 3,000
x + 6y = 1,800.
We can now eliminate x by multiplying the second equation
by –3 and adding:
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Example 5 – Solution
So,
cont’d
y = 2,400/16
= 150.
Substituting this into the equation x + 6y = 1,800 gives
x + 900 = 1,800,
and so
x = 900.
The solution is (x, y) = (900, 150). In other words, the
company should mix 900 quarts of Beginner’s juice and
150 quarts of Advanced juice.
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