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Algebra/Geometry Blend
Name
Unit #4: Systems of Linear Equations & Inequalities
Lesson 3: Solving Systems of Linear Equations Algebraically
Period
Date
[DAY #1]
Earlier we talked about how to find the solution to a system of equations by graphing each
equation. That is often time a LOT more work and not always the best choice. Who walks
around with graph paper or a graphing calculator?
You can also solve equations algebraically. In fact, there are 2 different methods for
solving equations algebraically. Today we are going to learn and explore how to solve
systems of equations using substitution.
Substitution
 To substitute is to
a variable with something of equal value.
Examples: Find the solution to the following system of equations algebraically
#1.
y = 2x
6x – y = 8
Steps
1) Substitute….. plug in the equation that has
the variable __________ into the other
equation.
2) Solve the equation for the first ______________
3) Substitute…… plug in the answer for the
variable that you just found into one of the
________________ equations and ____________ for the
other variable.
4) Write out your answer.
#2.
3x – 5y = 11
x = 3y + 1
#3.
y=x–4
2x + y = 5
#4.
x + 4y = 6
x = -y + 3
#5.
2x + y = 1
x = 23 + 4y
[DAY #2]
Solve the following systems of equations by substitution.
1.)
3x + 4y = 2
2.)
x=y+3
3.)
3x – y = 12
y = 2x – 7
4.)
2x – 5y = 14
y = 3x + 5
5x + y = 9
3x + 2y = -3
Try these…
1.)
3x + 2y = 16
x = 2y – 8
2.)
3x – 8y = 17
y = 2x – 7
3.)
4.)
3x + y = 13
5x + 4y = -4
5x – y = 8
y = 3x
[DAY #3]
In your own words, write what you think the word Elimination means?
Example #1: Solve this system of linear equations using the elimination method.
Let’s solve this system of equations by eliminating the ‘x’ variable. If you look at the ‘x’
terms, we have an ‘x’ and a ‘2x’. What will cancel out the ‘2x’ term?
.
We need to multiply the ‘x’ term by
to turn it into
so it will
cancel out with the ‘2x’ term. But when we do, we need to multiply EVERYTHING in that
equation by the same number.
ORIGINAL SYSTEM
2𝑥 + 𝑦 = 6
𝑥 − 3𝑦 = −11
NEW SYSTEM
SOLUTION
Once you have the NEW SYSTEM, add the two equations together. One of the variables
should cancel out. Solve the remaining equation for the variable. Substitute that variable
into EITHER of the original equations (YOUR CHOICE!!!) and solve. Write out your answer!
Now let’s solve this system of equations by eliminating the ‘y’ variable. If you look at the ‘y’
terms, we have an ‘y’ and a ‘-3y’. What will cancel out the ‘-3y’ term?
.
We need to multiply the ‘y’ term by
to turn it into
so it will cancel out
with the ‘-3y’ term. But when we do, we need to multiply EVERYTHING in that equation by
the same number.
ORIGINAL SYSTEM
2𝑥 + 𝑦 = 6
𝑥 − 3𝑦 = −11
NEW SYSTEM
SOLUTION
 When using Elimination you must _____________________________ the entire equation by
a constant to eliminate one of the
 Sometimes you might need to multiply BOTH equations by a
to eliminate the variable
Example #2: Solve this system of linear equations using the elimination method.
ORIGINAL SYSTEM
2𝑥 + 3𝑦 = 7
𝑥−𝑦 =1
NEW SYSTEM
SOLUTION
Example #3: Solve this system of linear equations using the elimination method.
ORIGINAL SYSTEM
2x – 5y = 18
7x + y = 26
NEW SYSTEM
SOLUTION
Try these…
Solve each system of equations by writing a new system that eliminates one of the
variables.
1.)
5x + y = 6
2.)
2x + 5y = -20
3x – 4y = 22
7x + 5y = 5
3.)
7x – 3y = 2
2x + 6y = - 20
4.)
9x + 5y = 5
2x + 10y = -30
[DAY #4]
Some times when solving a system of equations by elimination, you have to multiply
BOTH equations in order for something to cancel out.
1.)
3x – 5y = 29
2.)
8x – 3y = - 13
2x + 3y = -6
3x + 5y = - 11
3.)
3x + 4y = -10
2x + 5y = -2
4.)
6x – 4y = 38
5x – 3y = 31
Try these…
1.)
2x + 4y = -8
3x – 5y = 21
2.)
7x + 2y = 42
3x – 8y = - 44
3.)
4.)
3x + 2y = - 14
7x + 8y = - 11
6x – 3y = 0
2x + y = - 10
[DAY #5]
We have explored 3 different ways to solve a system of equations. There will be times you
MUST solve a system a particular way. Other times, it is up to you to choose the method.
But how do you know when to use which method?
 Solve by graphing only when
 Solve by substitution when
or
of the variables are alone.
 Solve by elimination when the
are “lined up”
1.)
Method:
5x + y = 15
3x – 2y = -4
2.)
Method:
4x + 3y = 5
6x + 5y = 9
3.)
Method:
y=x+7
4x – 2y = -24
4.)
Method:
5x – 6y = -9
10x + 3y = -33
Let’s look at some Regents questions that have appeared on this topic.
5.) A system of equations is given below
x + 2y = 5
2x + y = 4
Which system of equations does not have the same solution?
(1)
3x + 6y = 15
2x + y = 4
(2)
x + 2y = 5
6x + 3y = 12
(3)
4x + 8y = 20
2x + y = 4
(4)
x + 2y = 5
4x + 2y = 12
6.)
Which pair of equations could not be used to solve the following equations
for x and y?
4x + 2y = 22
-2x + 2y = -8
(1)
4x + 2y = 22
2x – 2y = 8
(2)
12x + 6y = 66
6x – 6y = 24
(3)
4x + 2y = 22
-4x + 4y = - 16
(4)
8x + 4y = 44
-8x + 8y = -8