3. 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.
4. Methods of Solving Systems
Algebraically
We will look at TWO methods to solve
systems algebraically:
1) Substitution
2) Elimination
5. 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.
6. Method 1: Substitution
Example:
Equation ‘a’: 3x + 4y = - 4
Equation ‘b’: x + 2y = 2
Isolate the ‘x’ in equation ‘b’:
x = - 2y + 2
7. Method 1: Substitution
Example, continued:
Equation ‘a’: 3x + 4y = - 4
Equation ‘b’: x + 2y = 2
Substitute the new expression,
x = - 2y + 2 for x into equation ‘a’:
3(- 2y + 2) + 4y = - 4
9. Method 1: Substitution
Example, continued:
Equation ‘a’: 3x + 4y = - 4
Equation ‘b’: 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).
11. Method 2: Elimination
Steps:
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.
12. Method 2: Elimination
Steps, continued:
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.
16. 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)
17. Method 2: Elimination
Example 2, continued:
Equation ‘a’: - 4(-9x + 6y = 0)
Equation ‘b’: 3(-12x + 8y = 0)
36x - 24y = 0
-36x + 24y = 0
0=0
What does this answer mean?
Is it true?
18. Method 2: Elimination
36x - 24y = 0
Example 2, continued:
-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.
Since 0 = 0 is TRUE, there are infinite solutions.
19. Solving Systems of Three
Equations Algebraically
1. When we have three equations in a
system, we can use the same two
methods to solve them algebraically
as with two equations.
2. Whether you use substitution or
elimination, you should begin by
numbering the equations!
20. Solving Systems of Three
Equations
Substitution Method
1. Choose one of the three equations
and isolate one of the variables.
2. Substitute the new expression into
each of the other two equations.
3. These two equations now have the
same two variables. Solve this 2 x 2
system as before.
4. Find the third variable by
substituting the two known values
into any equation.
21. Solving Systems of Three
Equations
Linear Combination Method
1. Choose two of the equations and eliminate
one variable as before.
2. Now choose one of the equations from step 1
and the other equation you didn’t use and
eliminate the same variable.
3. You should now have two equations (one
from step 1 and one from step 2) that you
can solve by elimination.
4. Find the third variable by substituting the
two known values into any equation.
22. -4x – 8y = 16
6x + 12y = -24
1. Solve the system of linear
equations using elimination
(linear combination).
