Download Class Notes 11-23-2010

Section 4.3
Solving Systems on Linear Equations by Addition
Learning objectives
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Use the addition method to solve a system of linear
equations
Vocabulary: elimination method, opposite
Solving systems of linear equations
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The first way we learned to solve a system of linear equations
was: graphing both lines and finding the intersection of the two
lines
The ordered pair point where the two lines intersected
represented the solution to the system.
Next, we learned the substitution method
If one equation is solved for one of the variables, that equation
could be substituted for that variable in the other equation, to
solve for the other variable.
Once a value for the ‘other’ variable was found in the ‘other
equation’, that value was then substituted into the first
equation to find the first variable value
Example of the substitution method
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Solve the system:
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Since the first equation has an ‘x’ in it, then solve for x:
x – 4y = -1  x = 4y – 1
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x – 4y = -1 and 6x – 3y = -6
Now, substitute that value for x into the other equation:
6x – 3y = -6  6(4y – 1) – 3y = -6  24y – 6 – 3y = -6
21y – 6 = -6  21y = 0  y = 0
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Finally, since y = 0, substitute that into the first equation:
x – 4(0) = -1  x = -1
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The solution to the system is the point (-1,0)
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There is a third way to solve systems of
equations
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In most cases, using addition will take less work than
substitution.
Solve the system: x + y = 5 and x – y = 11
We could solve for one variable and start substituting
Or, we can line these equations up together and add them
x + y =5
+ x - y = 11
---------------------2x + 0y = 16
So then x = 8
Then substitute for x:
8 + y = 5  y = -3
Solution: (8,-3)
You can line up two equations and add
them together as if they were numbers
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-x + 4y = 28
+ -6x - 4y = -56
--------------------------------------7x
= -28
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Thus, x = 4. Substitute
into the first equation
-(4) + 4y = 28
4y = 32  y = 8
Solution: (4,8)
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-3x + 2y = 10
+ 3x - 4y = -2
------------------------------------2y = 8
Thus, y = -4
Now substitute:
-3x + 2(-4) = 10
-3x – 8 = 10
so x = -6, Solution: (-6,-4)
Solve these systems of equations by
addition:
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2x + 6y = 14; -2x – 2y = 2
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3x – 3y = 3; -2x + 3y = 7
So far these involved two equations with an
easy ‘elimination method’
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In the example –x + 4y = 28 and -6x – 4y = -56
You will note that the first equation has a +4y term and
the second has a -4y term
When you add the two equations together, the y-terms
naturally cancel out
Not all systems of equations are set up that nicely for us,
however.
How about: x + 5y = 49 and -7x + 4y = -31?
Directly adding them together will give us -6x + 9y = 18
Sometimes, we must use multiplication on
one of the equations to allow for elimination
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Once again, the LCM (Least Common Multiple) from
Chapter R rears its ugly head!
In x + 5y = 49 and -7x + 4y = -31, we want to set it up so,
when we add the two equations together, one of the
variables is eliminated.
The only way we can do that is by multiplying one or
both of the equations.
We want to try to make our choice so we only have to
multiply one of the equations
We make that choice by: finding the LCM of both the xterms and the y-terms
What is the LCM of the x and y terms?
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In x + 5y = 49 and -7x + 4y = -31,
The x-terms are x and -7x which have an LCM of 7
The y-terms are 5y and 4y which have an LCM of 20
To achieve elimination of the x-terms, we only have to
multiply one of the equations (the first one) by 7
To achieve elimination of the y-terms, we have to multiply
both equations
Let’s do it the easy way, and try to eliminate the x - term
Solving the system
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x + 5y = 49 (x 7) 
-7x + 4y = -31
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7x + 35y = 343
+ -7x + 4y = -31
----------------------------------------39y = 312
y=8
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Now, substitute ‘8’ for y:
x + 5(8) = 49
x=9
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The solution to our system is (9,8)
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Solve: x + 3y = 2 & 4x + 2y = 18
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The LCM of the x-terms is 4, and of the y-terms, 6
We will eliminate the x-terms, since it requires only one
multiplication
x + 3y = 2 
4x + 12y = 8
4x + 2y = 18  - 4x + 2y = 18  Subtraction!
--------------------------10 y = -10,  y = -1
Now substitute -1 for y
x + 3(-1) = 2  x – 3 = 2  x = 5
The solution to this system: (5,-1)
Solve these systems of equations by
addition:
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-2x + 3y = 13; 4x - 2y = 3
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4x + 4y = 4; -6x +2y = -10
Sometimes, though, you have no choice but
to multiply both equations
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Take -2x – 7y = -6 and 5x – 3y = -26
The LCM of the x-terms is 10, and the LCM of the yterms is 21
To eliminate the x-terms, we will have to multiply the first
equation by 5 and the second equation by 2 before adding
To eliminate the y-terms, we will have to multiply the first
by 3 and the second by 7 before adding
At this point you can either:
- choose one variable to eliminate and go for it
- revert back to the substitution method from last
chapter
Solve -2x – 7y = -6 and 5x – 3y = -26
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I will choose to eliminate the x-terms
-2x – 7y = -6 (* 5) 
-10x - 35y = -30
5x – 3y = -26 (* 2)  + 10x - 6y = -52
--------------------------------41y = -82
y=2
Now, substitute the value for y into the first equation
-2x – 7(2) = -6 
-2x – 14 = -6  -2x = 8
So x = -4
The solution to this system is (-4,2)
One more time
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Solve 3x + 1/3y = 10 and 2x + 2/3y = 4
With fractions, the same LCM rule applies
The LCM of the x-terms is 6 and the y-terms is 2/3
Since I will only have to multiply one equation, I will
eliminate the y-term
3x + 1/3y = 10 (*2)  6x + 2/3y = 20
2x + 2/3y = 4 
- 2x + 2/3y = 4
------------------------4x
= 16; x = 4
Substitute for x: 3(4) + 1/3y = 10; 1/3y = -2; y = -6
The solution of the equation is (4,-6)
Solve the systems of equations
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5x + 8y = 1; 2x + 3y = 2
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3.5x + 0.3y = -18.7;
0.7x + 0.9y = -7.1
Some systems have no solution
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Solve –x – 2y = -4 and 5x + 10y = 8
Both of these equations have a slope of -1/2, so they are
parallel lines and thus have no solution
But I will multiply the first equation by 5, the LCM:
-5x – 10y = -20
+ 5x + 10y = 8
-------------------------------------0x + 0y = -12  0 = -12
This system has no solutions since 0 = -12 is false
Solve:
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4x – 6y = 1 and 20x – 30y = 3
One minute quiz
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What are the three ways we can find the solution to a
system of linear equations?