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Solving Systems
of Equations
3 Approaches
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Ms. Nong
Adapted from Mrs. N. Newman’s PPT
Method #1
POSSIBLE ANSWER:
Graphically
Method #2
Algebraically Using
Addition and/or
Subtraction
Method #3
Algebraically Using
Substitution
Answer: (x, y)
or (x, y, z)
Answer: No Solution
Answer: Identity


In order to solve a system of
equations graphically you typically
begin by making sure both
equations are in Slope-Intercept
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.
How to Use Graphs to solve Linear Systems.
Looking at the System Graphs:
•If the lines cross once, there
will be one solution.
•If the lines are parallel, there
will be no solutions.
•If the lines are the same, there
will be an infinite number of
solutions.
Check by substitute answers to equations:
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
becomes
6 x  14 y  24
Therefore
5 y  15
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 one 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
Lets look at another Systems solve by Substitution
y = 4x
3x + y = -21
Step 5: Check the solution in both equations.
y = 4x
-12 = 4(-3)
-12 = -12
3x + y = -21
3(-3) + (-12) = -21
-9 + (-12) = -21
-21= -21
This concludes my presentation on simultaneous
equations.
Please feel free to view it again at your leisure.
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