Objective
You will be able to:
Solve systems of equations using
elimination with addition or (subtraction)
and multiplication.
UNIT 6
Concept 3
Solving Systems of Equations
So far, we have solved systems using
graphing and substitution. These notes
show how to solve the system
algebraically using ELIMINATION with
addition and subtraction.
Elimination is easiest when the
equations are in standard form.
The Linear Combination Method
aka The Addition Method, aka The
Elimination Method.
Add (or subtract) a multiple of
one equation to (or from) the
other equation, in such a way
that either the x-terms or the y-
terms cancel out.
Then solve for x (or y,
whichever's left) and substitute
back to get the other coordinate.
Solve this system of equations
using the addition or
subtraction method.
When solving systems of linear
equations using elimination, you
will sometimes need to multiply
one or both equations by a factor
in order to get the same
coefficients for a variable.
Best time to use this method is
when you have two coefficients
that the same
Steps in Elimination
1. Line up the two equations
using standard form
(Ax + By = C).
2. GOAL: The coefficients of
the same variable in both
equations should have the
same value but opposite
signs.
3. If this doesn’t exist, multiply
one or both of the equations
by a number that will make
the same variable
coefficients opposite
values.
4. Add the two equations
(like terms).
5. The variable with
opposite coefficients
should be eliminated.
6. Solve for the
remaining variable.
7. Substitute that
solution into either of
the two equations to
solve for the other
variable.
Color Coded Steps in solving a system of
equations by elimination.
Step 1: Put the equations in Standard Form.
Step 2: Determine which
variable to eliminate.
Step 3: Add or subtract the
equations.
Step 4: Plug back in to find
the other variable.
Step 5: Check your
solution.
Standard Form: Ax + By = C
Look for variables that have the
same coefficient.
Solve for the variable.
Substitute the value of the variable
into the equation.
Substitute your ordered pair into
BOTH equations.
1) Solve the system using elimination.
x + y = 5
3x – y = 7
Step 1: Put the equations in
Standard Form.
Step 2: Determine which
variable to eliminate.
They already are!
The y’s have the same
coefficient.
Step 3: Add or subtract the
equations.
Add to eliminate y.
x + y = 5
(+) 3x – y = 7
4x = 12
x = 3
Add into your
spirals
1) Solve the system using elimination.
Step 4: Plug back in to find
the other variable.
x + y = 5
(3) + y = 5
y = 2
Step 5: Check your
solution.
(3, 2)
(3) + (2) = 5
3(3) - (2) = 7
The solution is (3, 2). What do you think the answer
would be if you solved using substitution?
x + y = 5
3x – y = 7
2) Solve the system using elimination.
4x + y = 7
4x – 2y = -2
Step 1: Put the equations in
Standard Form. They already are!
Step 2: Determine which
variable to eliminate.
The x’s have the same
coefficient.
Step 3: Add or subtract the
equations.
Subtract to eliminate x.
4x + y = 7
(-) 4x – 2y = -2
3y = 9
y = 3
Remember to
“keep-change-
change”
Add into your
spirals
2) Solve the system using elimination.
Step 4: Plug back in to find
the other variable.
4x + y = 7
4x + (3) = 7
4x = 4
x = 1
Step 5: Check your
solution.
(1, 3)
4(1) + (3) = 7
4(1) - 2(3) = -2
4x + y = 7
4x – 2y = -2
3) Solve the system using elimination.
y = 7 – 2x
4x + y = 5
Step 1: Put the equations in
Standard Form.
2x + y = 7
4x + y = 5
Step 2: Determine which
variable to eliminate.
The y’s have the same
coefficient.
Step 3: Add or subtract the
equations.
Subtract to eliminate y.
2x + y = 7
(-) 4x + y = 5
-2x = 2
x = -1
Add into your
spirals
2) Solve the system using elimination.
Step 4: Plug back in to find
the other variable.
y = 7 – 2x
y = 7 – 2(-1)
y = 9
Step 5: Check your
solution.
(-1, 9)
(9) = 7 – 2(-1)
4(-1) + (9) = 5
y = 7 – 2x
4x + y = 5
Example 1 (In note packet pg 10)
Step 1: Put the equations in
Standard Form.
Step 2: Determine which
variable to eliminate.
Step 3: Add or subtract the
equations.
x - 2y = 14
x + 3y = 9
Both equations are already in
Standard Form.
We will eliminate the variable “x”
Since they both have same coefficient.
Subtraction is easiest, just remember this
Is like multiplying the entire equation by
-1.
x - 2y = 14
-( x + 3y = 9)
-5y = 5
y = -1
Example 1 continued…
Step 4: Plug back in to find
the other variable.
Step 5: Check your
solution.
Since y = -1, we will substitute in
x - 2y = 14
x- 2 (-1) = 14 Therefore
x+2=14 my solution
x=12 is ( 12, -1)
Substitute ( 12, -1) into:
x - 2y = 14
12 – 2(-1) = 14
12+2 = 14
14 = 14
x - 2y = 14
x + 3y = 9
Example 2 (In note packet pg 11)
Step 1: Put the equations in
Standard Form.
Step 2: Determine which
variable to eliminate.
Step 3: Add or subtract the
equations.
4x + 3y = -1 5x + 4y = 1
Both equations are already in
Standard Form.
So for this one we need to multiply
each equation to create same
Coefficients. So we will multiply
EQ1 by 4 and EQ2 by -3,
then add to eliminate y.
4(4x + 3y = -1) -3(5x + 4y = 1)
16x + 12y = -4) -15x - 12y = -3) x = -7
Example 2 continued…
Step 4: Plug back in to find
the other variable.
Step 5: Check your
solution.
4x + 3y = -1
4(-7) + 3y = -1 Therefore
-28 +3y = -1 my solution is
3y = 27
y = 9 (-7, 9)
4x + 3y = -1 5x + 4y = 1
4(-7) +3(9) = -1 (-7) + 4(9) = 1
-28 + 27 = -1 -35 + 36 = 1
-1 = -1 (check!) 1 = 1
4x + 3y = -1 5x + 4y = 1
The third example we will do together in class.
Example 1:
Equation ‘a’: 2x - 4y = 13
Equation ‘b’: 4x - 5y = 8
Multiply equation ‘a’ by –2 to
eliminate the x’s:
Equation ‘a’: -2(2x - 4y = 13)
Equation ‘b’: 4x - 5y = 8
EXTRA EXAMPLES
EXTRA EXAMPLES
Extra Example 1, continued:
Equation ‘a’: -2(2x - 4y = 13) ------> -4x + 8y = -26
Equation ‘b’: 4x - 5y = 8 ------> 4x - 5y = 8
Add the equations (the x’s are
eliminated): -4x + 8y = -26
4x - 5y = 8
3y = -18
y = -6
EXTRA EXAMPLES
-22
4-11
2
Example 1, continued:
Equation ‘a’: -2(2x - 4y = 13) ------> -4x + 8y = -26
Equation ‘b’: 4x - 5y = 8 ------> 4x - 5y = 8
Substitute y = -6 into either equation:
-11
2
4x - 5(-6) = 8
4x + 30 = 8
4x = -22
x =
x =
Solution: ( , -6)
EXTRA EXAMPLES
Extra Example 2: Equation ‘a’: -9x + 6y = 0
Equation ‘b’: -12x + 8y = 0
Equation ‘a’: - 4(-9x + 6y = 0)
Equation ‘b’: 3(-12x + 8y = 0)
Multiply equation ‘a’ by –4 and equation ‘b’
by 3 to eliminate the x’s:
36x - 24y = 0
-36x + 24y = 0
0 = 0
Equation ‘a’: - 4(-9x + 6y = 0)
Equation ‘b’: 3(-12x + 8y = 0)
What does this answer mean?
Is it true?
Example 2 continued…
EXTRA EXAMPLES
36x - 24y = 0
-36x + 24y = 0
0 = 0
Example 2, continued:
When both variables are eliminated,
if the statement is TRUE (like 0 = 0), then they are the same lines and there are infinite solutions.
if the statement is FALSE (like 0 = 1), then they are parallel lines and there is no solution.
36x - 24y = 0
-36x + 24y = 0
0 = 0
Since 0 = 0 is TRUE, there are infinite solutions.
EXTRA EXAMPLES
Summary
When solving systems of linear equations using elimination, you
will sometimes need to multiply one or both equations by a factor
in order to get the same coefficients for a variable. This process is
very similar to getting a common denominator for fractions.
So to summarize, this is the most complex method of solving systems of
of equations. Although you just need to be sure to follow the steps to
ensure success.
Remember:
ASSIGNMENT
• Do your summary from the guided questions online.
• Do practice problems in note packet on page 12
• Do practice quiz on page 13.
• Section 6.3 pages 347-34
Do Problems 1-10; 33 & 34, and 43-45 9