Download Lesson 5.3 Notes

Name_______________________________________Date______________Period____
Lesson 5.3
“Solving Systems of Equations: Elimination”
Remember that the solution for a system MUST SATISFY both linear equations!
The first method of solving a system of equations you learned was called
_________________.
Graphing works nicely when you have whole numbers to work with…
The second method focused more on an algebraic approached, called _______________
This is when you have a variable that is isolated and set equal to the rest of the
equation, or you can simply transform a variable to be isolated.
The final method of solving a system of equations, is called
________________________________.
Elimination is best used when both of your variables have a number attached to them.
You will then need to find factors, that when multiplied, will ELIMINATE one of the
variables. Let’s take a look at what that looks like:
The steps to solve a system of equations using ELIMINATION, are as follow:
1. Make sure the equations line up vertically. Make sure your “x” variables line up,
your “y” variables, and your equal sign/constant.
2. Look to see if either of the variables will naturally cancel out. It would look like
the following, where you have the opposite signs with the same number:
a. 3x – 5y = 16
2x + 5y = 4
b. You can see here that the -5 and +5 would naturally cancel out
3. If neither of your variables naturally cancel out, you will need to multiply them
by number to make them cancel out. It would look like the following:
a. 4x – 2y = 8
2x + 3y = 4
You can see here that the 4 and 2 do not cancel, nor do the -2 and 3.
However, since the -2 and 3 are opposite signs, I would start here.
What can we multiply both of them by to equal a like multiple? What
would that multiple be?
b. SIX! Six is a common multiple for both 2 and 3. Therefore, I am going to
multiply the top equation by 3, and the bottom equation by 2.
3 (4x – 2y = 8)
2 (2x + 3y = 4)
c. Now simplify:
12x – 6y = 24
4x + 6y = 8
4. Now you can see that the “y” variables will cancel out, leaving us with:
12x = 24
4x = 8
5. Add each column vertically:
16 x = 32
Solve for x:
x=2
6. Plug in “x” to solve for “y” (pick the easiest ORIGINAL equation)
4x – 2y = 8
4(2) – 2y = 8
8 – 2y = 8
-2y = 0
y=0
7. Your solution is an ordered pair of (2, 0)
Sounds complicated, right? Let’s try a couple more examples!
Using ELIMINATION, find the solution of the following system of equations.
-2x + 3y = 4
2x – y = -8
Ask yourself; do any of your variables naturally cancel out?
YES – the x variables! So, we are left with:
3y = 4
-y = -8
Add each column (the y – variables and the constants) vertically:
2y = -4
Solve for y:
y = -2
Pick an equation, plug “y = -2” in and solve for “x”:
-2x + 3y = 4
-2x + 3(-2) = 4
-2x – 6 = 4
-2x = 10
x = -5
The solution to the system is (-5, -2)
Using ELIMINATION, find the solution to the following system:
4x – 3y = 8
5x – 2y = -11
Ask yourself; do any of your variables naturally cancel out?
NO – so we will need to multiply each equation by a number that creates the same
multiple with different signs. Neither of the variables have opposite signs, so let’s just
start by eliminating “y”, solving for “x”:
2 (4x – 3y = 8)
-3 (5x – 2y = -11)
Simplify:
8x – 6y = 16
-15x + 6y = 33
“y” variables cancel out, and you are left with:
8x = 16
-15x = 33
Add each column (the y – variables and the constants) vertically:
-7x = 49
x = -7
Pick an equation, plug “x = -7” in and solve for “y”:
4x – 3y = 8
4 (-7) – 3y = 8
-28 - 3y = 8
-3y = 36
y = -12
The solution to the system is (-7, -12)
----------------Let’s try it together!
5x + 6y = 50
3x + 2y = -2
x – 6y = -26
-6x – 5y = 2
PRACTICE TIME!! 