1. Lesson 4-2: Solving Systems – Substitution & Linear Combinations
Objectives:
Students will:
Solve systems of equations using substitution and linear combinations

2. Substitution (means replace)
Solve one equation for one of the variables (find an easy one!)
Substitute what that variable equals into the other equation.
Solve for the remaining variable.
Substitute the value of this variable back into step 1
Solve for the other variable.
Check that the two values solves both equations

3. (-1 - x) The solution is (4.5, -5.5) Example 1 Solve the system
5x + 3y = 6
y = x
5x = 6
5x x = 6
2x – 3 = 6
x =
(-1 - x)
y = -1 – x
y = -1 -
y =
Wow step 1 is done!
The solution is (4.5, -5.5)

4. Example 2: x + 2y = 6 4x + 3y = 4 x = 6 - 2y x = 6 - 2y x = 6 -2(4)
This one was easy to solve for x.
Look for variable with no coefficient
x + 2y = 6
4x + 3y = 4
x = 6 - 2y
x = 6 - 2y
x = 6 -2(4)
x = -2
+ 3y = 4
24 - 8y + 3y = 4
-5y = -20
y = 4
(6 - 2y) x
The solution is (-2, 4)

5. Example 3
5x + 3y = 17
-5x + 2y = 3
Solving for either variable is a challenge here!!!
There has to be an easier way!

6. Linear Combination (ELIMINATION)
Set up both equations in Standard Form : Ax + By = C
Obtain opposite x or y terms. Multiply (one or both) equations by a number if you have to make opposites.
Add the equations vertically – x’s or y’s will eliminate
Solve for the remaining variable.
Substitute the value of this variable back into one of the original equations
Solve for the other variable.
Check that the two values solves both equations

7. 5x +3(4) =17 5x +12 =17 x = 1 Example 3 Use elimination 5x + 3y = 17
This is an easy one x’s are already opposite
The solution is (1,4)
This time let’s make the y’s opposite. We need to change both equations
Example 4
6x + 2y = -16
-12x - 5y = 31
( ) x +10y = -80
( ) x - 10y = 62 +
Finishing this one now will be a piece of cake!

8. ( ) 6 ( ) 12 Example 4 What would make this easier… I remember!
Get rid of fractions by multiplying by LCD
Example 4
( ) 6
( ) 12
Now you can do it!
Give it a try!!!