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Good Morning,
We are moving on to chapter 3.
 If there is time today I will show you
your test score you can not have them
back as I still have several folks that
have not finished
 Quiz Tuesday (9/12) on solving
algebraically
 Chapter 3 homework will be due 9/19

SYSTEMS OF LINEAR
EQUATIONS
Solving Linear Systems Algebraically
Solving Systems of Equations
Algebraically
1. When you graph, sometimes you cannot find
the exact point of intersection. We can use
algebra to find the exact point.
2. Also, we do not need to put every equation in
slope-intercept form in order to determine if
the lines are parallel or the same line.
Algebraic methods will give us the same
information.
Methods of Solving Systems
Algebraically
We will look at TWO methods to solve
systems algebraically:
1) Substitution
2) Elimination
Method 1: Substitution
Steps:
1. Choose one of the two equations
and isolate one of the variables.
2. Substitute the new expression into
the other equation for the variable.
3. Solve for the remaining variable.
4. Substitute the solution into the
other equation to get the solution
to the second variable.
Example #1:
y = 4x
3x + y = -21
Step 1: Solve one equation for one variable.
y = 4x
(This equation is already solved for y.)
Step 2: Substitute the expression from step one into
the other equation.
3x + y = -21
3x + 4x = -21
Step 3: Simplify and solve the equation.
7x = -21
x = -3
y = 4x
3x + y = -21
Step 4: Substitute back into either original
equation to find the value of the other
variable.
3x + y = -21
3(-3) + y = -21
-9 + y = -21
y = -12
Solution to the system is (-3, -12).
y = 4x
3x + y = -21
Step 5: Check the solution in both equations.
Solution to the system is (-3,-12).
y = 4x
-12 = 4(-3)
-12 = -12
3x + y = -21
3(-3) + (-12) = -21
-9 + (-12) = -21
-21= -21
Example #2:
x + y = 10
5x – y = 2
Step 1: Solve one equation for one variable.
x + y = 10
y = -x +10
Step 2: Substitute the expression from step one into
the other equation.
5x - y = 2
5x -(-x +10) = 2
x + y = 10
5x – y = 2
Step 3: Simplify and solve the equation.
5x -(-x + 10) = 2
5x + x -10 = 2
6x -10 = 2
6x = 12
x=2
x + y = 10
5x – y = 2
Step 4: Substitute back into either original
equation to find the value of the other
variable.
x + y = 10
2 + y = 10
y=8
Solution to the system is (2,8).
x + y = 10
5x – y = 2
Step 5: Check the solution in both equations.
Solution to the system is (2, 8).
x + y =10
2 + 8 =10
10 =10
5x – y = 2
5(2) - (8) = 2
10 – 8 = 2
2=2
Example # 3 Substitution
Example:
Equation ‘a’:
Equation ‘b’:
3x + 4y = - 4
x + 2y = 2
Isolate the ‘x’ in equation ‘b’:
x = - 2y + 2
Example, continued:
Equation ‘a’:
Equation ‘b’:
3x + 4y = - 4
x + 2y = 2
Substitute the new expression,
x = - 2y + 2 for x into equation ‘a’:
3(- 2y + 2) + 4y = - 4
Example, continued:
Equation ‘a’:
Equation ‘b’:
3x + 4y = - 4
x + 2y = 2
Solve the new equation:
3(- 2y + 2) + 4y = - 4
- 6y + 6 + 4y = - 4
- 2y + 6 = - 4
- 2y = - 10
y= 5
Example, continued:
Equation ‘a’:
Equation ‘b’:
3x + 4y = - 4
x + 2y = 2
Substitute y = 5 into either equation ‘a’ or ‘b’:
x + 2 (5) = 2
x + 10 = 2
x=-8
The solution is (-8, 5).
Solve by substitution:
1. y  2x  2
2x  3y  10
2. 2a  3b  7

2a  b  5
Solving Systems of Equations
using Elimination
Steps:
1. Place both equations in Standard Form, Ax + By = C.
2. Determine which variable to eliminate with Addition
or Subtraction.
3. Solve for the variable left.
4. Go back and use the found variable in step 3 to find
second variable.
5. Check the solution in both equations of the system.
EXAMPLE #1:
5x + 3y = 11
5x = 2y + 1
STEP1: Write both equations in Ax + By = C
form.
5x + 3y =11
5x - 2y =1
STEP 2:
Use subtraction to eliminate 5x.
5x + 3y =11
5x + 3y = 11
-(5x - 2y =1)
-5x + 2y = -1
Note: the (-) is distributed.
STEP 3:
Solve for the variable.
5x + 3y =11
-5x + 2y = -1
5y =10
y=2
5x + 3y = 11
5x = 2y + 1
STEP 4:
Solve for the other variable by substituting
into either equation.
5x + 3y =11
5x + 3(2) =11
5x + 6 =11
5x = 5
x=1
The solution to the system is (1,2).
5x + 3y= 11
5x = 2y + 1
Step 5:
Check the solution in both equations.
The solution to the system is (1,2).
5x + 3y = 11
5(1) + 3(2) =11
5 + 6 =11
11=11
5x = 2y + 1
5(1) = 2(2) + 1
5=4+1
5=5
Solving Systems of Equations
using Elimination
Steps:
1. Place both equations in Standard Form, Ax + By = C.
2. Determine which variable to eliminate with Addition
or Subtraction.
3. Solve for the remaining variable.
4. Go back and use the variable found in step 3 to find
the second variable.
5. Check the solution in both equations of the system.
Example #2:
x + y = 10
5x – y = 2
Step 1: The equations are already in standard
form:
x + y = 10
5x – y = 2
Step 2: Adding the equations will eliminate y.
x + y = 10
x + y = 10
+(5x – y = 2)
+5x – y = +2
Step 3:
Solve for the variable.
x + y = 10
+5x – y = +2
6x = 12
x=2
x + y = 10
5x – y = 2
Step 4:
Solve for the other variable by
substituting into either equation.
x + y = 10
2 + y = 10
y=8
Solution to the system is (2,8).
x + y = 10
5x – y = 2
Step 5:
Check the solution in both equations.
Solution to the system is (2,8).
x + y =10
2 + 8 =10
10=10
5x – y =2
5(2) - (8) =2
10 – 8 =2
2=2
Method 2: Elimination
Example:
Equation ‘a’:
Equation ‘b’:
2x - 4y = 13
4x - 5y = 8
Multiply equation ‘a’ by –2 to eliminate
the x’s:
Equation ‘a’:
Equation ‘b’:
-2(2x - 4y = 13)
4x - 5y = 8
Method 2: Elimination
Example, continued:
Equation ‘a’:
Equation ‘b’:
-2(2x - 4y = 13) ------> -4x + 8y = -26
4x - 5y = 8
------> 4x - 5y = 8
Add the equations (the x’s are eliminated):
-4x + 8y = -26
4x - 5y = 8
3y = -18
y = -6
Method 2: Elimination
Example, continued:
Equation ‘a’: -2(2x - 4y = 13) ------> -4x + 8y = -26
Equation ‘b’: 4x - 5y = 8
------> 4x - 5y = 8
Substitute y = -6 into either equation:
4x - 5(-6) = 8
4x + 30 = 8
4x = -22
-22
x= 4
-11
x= 2
-11
2
Solution: ( , -6)
Method 2: Elimination
Example 2:
Equation ‘a’: -9x + 6y = 0
Equation ‘b’: -12x + 8y = 0
Multiply equation ‘a’ by –4 and
equation ‘b’ by 3 to eliminate the x’s:
Equation ‘a’: - 4(-9x + 6y = 0)
Equation ‘b’: 3(-12x + 8y = 0)
Method 2: Elimination
Example 2, continued:
Equation ‘a’:
Equation ‘b’:
- 4(-9x + 6y = 0)
3(-12x + 8y = 0)
36x - 24y = 0
-36x + 24y = 0
0=0
What does this answer mean?
Is it true?
Method 2: Elimination
Example 2, continued:
36x - 24y = 0
-36x + 24y = 0
0=0
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.
Method 2: Elimination
Example 2, continued:
36x - 24y = 0
-36x + 24y = 0
0=0
Since 0 = 0 is TRUE, there
are infinite solutions.
NOW solve these using elimination:
1.
2.
2x + 4y =1
x - 4y =5
2x – y =6
x+y=3