Transcript Solving Systems of Equations - Schenectady City School

Solving Systems
of Equations
3 Approaches
Click here to begin
Mrs. N. Newman
Method #1
Door #1
Graphically
Method #2
Algebraically Using
Addition and/or
Subtraction
Door #2
Method #3
Algebraically Using
Substitution
Door #3
In order to solve a system of
equations graphically you typically
begin by making sure both
equations are in standard form.
y  mx  b
Where m is the slope and b is the y-intercept.
Examples:
y = 3x- 4
Slope is 3 and y-intercept is - 4.
y = -2x +6
Slope is -2 and y-intercept is 6.
Graph the line by locating the appropriate
intercept, this your first coordinate. Then move to your next
coordinate using your slope.
Use this same process and graph the second line.
Once both lines have been graphed locate the point of
intersection for the lines. This point is your solution set.
In this example the solution set is [2,2].
In order to solve a system of equations algebraically
using addition first you must be sure that both
equation are in the same chronological order.
Example
:
Could be
y  x4
x y 2
yx4
yx2
Now select which of the two variables you want to eliminate.
For the example below I decided to remove x.
yx4
yx2
The reason I chose to eliminate x is because they
are the additive inverse of each other. That means
they will cancel when added together.
Now add the two equations together.
yx4
yx2
Your total is:
therefore
2y  6
y3
Now substitute the known value into either one of the original equations.
I decided to substitute 3 in for y in the second equation.
3 x  2
x  1
Now state your solution set always remembering to
do so in alphabetical order.
[-1,3]
Lets suppose for a moment that the
equations are in the same sequential
order. However, you notice that
neither coefficients are additive
inverses of the other.
2x  3y  3
3 x  7 y  12
Identify the least common multiple
of the coefficient you chose to
eliminate. So, the LCM of 2 and 3
in this example would be 6.
Multiply one or both equations
by their respective multiples. Be
sure to choose numbers that will
result in additive inverses.
 3(2 x  3 y  3) becomes
2(3x  7 y  12)
 6 x  9 y  9
6 x  14 y  24
Now add the two equations together.
 6 x  9 y  9
6 x  14 y  24
becomes
5 y  15
Therefore
y 3
Now substitute the known value into either one of the original
equations.
y3
2 x  3(3)  3
2x  9  3
2 x  6
x  3
Now state your solution set always remembering
to do so in alphabetical order.
[-3,3]
In order to solve a system equations algebraically using substitution you must have
on variable isolated in one of the equations. In other words you will need to solve
for y in terms of x or solve for x in terms of y.
In this example it has been done
for you in the first equation.
y  x4
x y 2
Now lets suppose for a moment that you are given a set of equations
like this..
2x  3y  3
3 x  7 y  12
Choosing to isolate y in the first equation the result is :
2
y   x 1
3
Now substitute what y equals into the second equation.
y  x4
x y 2
becomes
Better know as
Therefore
x x4 2
2x  4  2
2 x  2
x  1
This concludes my presentation on simultaneous equations.
Please feel free to view it again at your leisure.