Algebra (Period 1)

SOLVING SYSTEMS OF EQUATIONS 8-1 TO 8-3

(2 equations with 2 variables)
You cannot solve an equation with 2 variables - you can find multiple coordinates that work
TO SOLVE MEANS THE
ONE COORDINATE THAT WORKS FOR BOTH EQUATIONS

There are 3 ways to find that point:
1. Graph both equations: Where the 2 lines intersect is the solution
2. Substitution method: Solve one of the equations for either x or y and plug in to the other equation
3. Addition method: Eliminate one of the variables by multiplying the equations by that magical number that will make one of the variables the ADDITIVE INVERSE of the other

Example solved all 3 ways:
Find the solution to the following system:
2x + 3y = 8 and 5x + 2y = -2

1. GRAPH BOTH LINES: Put both in y = mx + b form and graph
Read the intersection point [you should get (-2, 4)]

2. SUBSTITUTION:
Isolate whatever variable seems easiest
I will isolate y in the second equation: 2y = -5x - 2
y = -5/2 x - 1
Plug this -5/2 x - 1 where y is in the other equation
2x + 3(-5/2 x - 1) = 8
2x - 15/2 x - 3 = 8
4/2 x - 15/2 x - 3 = 8
-11/2 x - 3 = 8
-11/2 x = 11
-2/11(-11/2 x) = 11(-2/11)
x = -2
Plug into whichever equation is easiest to find y

3. ADDITION:
Multiply each equation so that one variable will "drop out" (additive inverse)
I will eliminate the x, but could eliminate the y if I wanted to
5(2x + 3y) = (8)5
-2(5x + 2y) = (-2)-2

10x + 15y = 40
-10x - 4y = 4

ADD TO ELIMINATE THE x term:
11y = 44
y = 4
Plug into whichever equation is easiest to find x

NOTICE THAT FOR ALL 3 METHODS, THE SOLUTION IS THE SAME!
THEREFORE, USE WHATEVER METHOD SEEMS EASIEST!!!