Chapter 6-1, 6-2, 6-3
An overview of all
methods
Solving Systems of
Equations by Graphing
Before we stopped and reviewed fraction operations last week ( I didn’t) we had just finished graphing linear functions with two variables in three different forms:
Before we stopped and reviewed fraction operations last week ( I didn’t) we had just finished graphing linear functions with two variables in three different forms:
Slope-intercept:
Graphing our home base on the y-axis, then counting slope (rise/run) to another
point
Standard: Making
an x-y table to find the intercepts and graphing those
Point-slope:
Graphing the point in the formula (flip the signs!) and then counting slope to
another point
Today, we’ll have
a SYSTEM of linear functions and will find the solution (coordinate) to where
they INTERSECT.
Systems are 2 or
more linear functions (lines)
TO SOLVE MEANS TO FIND THE ONE COORDINATE THAT WORKS FOR BOTH FUNCTIONS
THERE ARE 3
POSSIBILITIES FOR THIS:
1. The lines
intersect at ONE POINT. We call this a consistent system that is independent.
2. The lines are
COLLINEAR, meaning that they intersect at INFINITELY MANY POINTS. We call this
consistent and dependent.
3. The lines are
PARALLEL, meaning that they NEVER INTERSECT and therefore there is NO POSSIBLE
SOLUTIONS. We call these lines inconsistent.
There are 3 ways to find that point:
(Chapter 3-7) 1. Graph both equations: Where the 2 lines intersect is the solution
(Chapter 3-8) 2. Substitution method: Solve one of the equations for either x or y and plug in to the other equation
There are 3 ways to find that point:
(Chapter 3-7) 1. Graph both equations: Where the 2 lines intersect is the solution
(Chapter 3-8) 2. Substitution method: Solve one of the equations for either x or y and plug in to the other equation
(NOT GIVEN IN YOUR
BOOK) 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
I will be doing the following example solved all 3 ways today.
I will be doing the following example solved all 3 ways today.
Find the solution to the following system…Find the COORDINATE where the lines intersect:
2x + y = -3 and 2x
- y = -5
1. GRAPH BOTH LINES: Put both in y = mx + b form and graph
Read the intersection point....You should get (-2, 1)
1. GRAPH BOTH LINES: Put both in y = mx + b form and graph
Read the intersection point....You should get (-2, 1)
We’ll also use the
GRAPHING CALCULATORS to find this intersection!
What’s the
drawback of this method? It takes time…Many times the intersection’s coordinate
is not an integer.
Solving systems of
linear equations algebraically (without graphing)
There are two
methods: SUBSTITUTION (Chapter 3-8) and ADDITION (not in your book)
2. SUBSTITUTION:
Isolate whatever variable seems easiest.
I will isolate y in the first equation: 2x + y = -3
I will isolate y in the first equation: 2x + y = -3
y = -2x – 3 (after
you subtract 2x from both sides)
Now plug in (-2x –
3) for y in the other equation:
2x - y = -5
2x – (-2x – 3) = -5
2x – (-2x – 3) = -5
2x + 2x + 3 = -5
4x + 3 = -5
4x = -8
x = -2
Now plug in -2 for
x in whichever equation looks easier to find y.
I think the first
equation looks easier:
2x + y = -3
2(-2) + y = -3
-4 + y = -3
Y = 1
So the coordinate
of the intersection is (-2, 1)
This is the same
as we found when we graphed.
To really be sure
you didn’t make a silly mistake, you should plug in the coordinate in the OTHER
equation:
2x - y = -5
2(-2) – (1) = -5????
2(-2) – (1) = -5????
-4-1=-5 YES!
When would it be
best to solve this way?
When one of the
equations is already solved for one of the variables, but you can always
isolate one of the variables yourself with equation balancing.
A lot of word
problems are easier to solve with substitution.
3. ADDITION:
Multiply each equation so that one variable will "drop out" (additive
inverses….YAY!)
For the problem above, I will eliminate the y because the two y’s are already Additive Inverses, but I could eliminate the x if I wanted to!
For the problem above, I will eliminate the y because the two y’s are already Additive Inverses, but I could eliminate the x if I wanted to!
This time you
“stack” the equations:
2x
+ y = -3
+ 2x
- y = -5
-----------------
-----------------
4x = -8
x = -2
Plug into
whichever equation is easiest to find y as we did in the Substitution method.
When is it best to
use this method?
If no variable is
already isolated.
I tend to
use this method the most ;)
NOTICE THAT FOR ALL 3 METHODS, THE SOLUTION IS THE SAME!
THEREFORE, USE WHATEVER METHOD SEEMS EASIEST!!!
NOTICE THAT FOR ALL 3 METHODS, THE SOLUTION IS THE SAME!
THEREFORE, USE WHATEVER METHOD SEEMS EASIEST!!!
