Download Chapter 3.2

Solving Systems of
Equations Algebraically
Chapter 3.2
Alternatives to Graphing

Sometimes, graphing systems is not the best
way to go
–
–
–

Lines don’t intersect at a discernable point
Don’t have enough room to graph
Etc.
Alternate methods of solving systems
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–
Substitution
Elimination
Substitution

Substitution:
–
One equation is solved for one variable in terms
of the other. This expression can be substituted
for the variable in the other equation.
Example 1

Solve either equation
for one variable

Substitute this
expression into the
other equation
– Now, there should
only be one
variable

After solving for one
variable, plug it into
either equation to
solve for other
variable
x + 4y = 26
x - 5y = -10
x = 5y – 10
x + 4y = 26
x + 4(4) = 26
x + 16 = 26
x = 10
5y – 10 + 4y = 26
9y – 10 = 26
9y = 36
y=4
Solution:
(10, 4)
Example 2

Solve either equation
for one variable

Substitute this
expression into the
other equation
– Now, there should
only be one
variable

After solving for one
variable, plug it into
either equation to
solve for other
variable
2x + y = 4
3x + 2y = 1
y = -2x + 4
2x + y = 4
2(7) + y = 4
14 + y = 4
y = -10
3x + 2(-2x + 4) = 1
3x - 4x + 8 = 1
-x + 8 = 1
-x = -7
x=7
Solution:
(7, -10)
Elimination

Elimination:
–
Eliminate one of the variables by adding or
subtracting the two equations together
Example 3

Instead of using
Substitution, you
can subtract one
equation from the
other
–

Subtract the 2nd from
the 1st
After solving for one
variable, plug it into
either equation to
find the other
x + 2y = 10
x+y=6
0+y=4
y=4
x+y=6
x+4=6
x=2
Solution:
(2, 4)
Example 4


Instead of using
Substitution, you
can add one
equation to the
other
2x + y = 5
3x - y = 20
5x + 0 = 25
After solving for one x = 5
variable, plug it into
either equation to
find the other
2x + y = 5
2(5) + y = 5
10 + y = 5
y = -5
Solution:
(5, -5)
Elimination with Multiplication

When one variable cannot be easily
eliminated using simple addition or
subtraction, multiply one or both equations
by constants so that a variable CAN be
eliminated.
Example 5

Simple addition or
subtraction isn’t
going to help here.

Decide which
variable to eliminate x5
–
Let’s do x

Now subtract 2nd
from 1st

Use one variable to
solve for the other
10x + 15y = 60
10x - 4y = 22
2x + 3y = 12
2x + 3(2) = 12
2x + 6 = 12
2x = 6
x=3
0 + 19y = 38
y=2
Solution:
(3, 2)
2x + 3y = 12
5x - 2y = 11
x2
Example 6

Simple addition or
subtraction isn’t
going to help here.

Decide which
variable to eliminate x2
–
Let’s do h

Now add together

Use one variable to
solve for the other
2g + h = 6
3g - 2h = 16
4g + 2h = 12
3g - 2h = 16
7g + 0 = 28
g=4
2g + h = 6
2(4) + h = 6
8+h=6
h = -2
Solution:
(4, -2)
Inconsistent and Dependent Systems

If you add or subtract two equations in a system and the result
is an equation that is never true, then the system is
inconsistent and it has no solution.
–
Examples



1=2
-1 = 1
If the result is an equation that is always true, then the system
is dependent and has infinitely many solutions.
–
Examples


1=1
9=9
Example 7 (substitution)

Solve either
equation for one
variable
x + 3y = 8
1/3 x + y = 9
x = -3y + 8

Substitute this
expression into
the other equation
– Now, there
should only be
one variable
1/3(-3y + 8) + y = 9
-y + 8/3 + y = 9
8/3 = 9
No Solutions!!
Not True
Example 8 (substitution)

Solve either
equation for one
variable
2a - 4b = 6
-a + 2b = -3
a = 2b + 3

Substitute this
expression into
the other equation
– Now, there
should only be
one variable
Infinitely Many
Solutions!!
2(2b + 3) - 4b = 6
4b + 6 - 4b = 6
6=6
Always True
Example 9 (elimination)


Simple addition or
subtraction isn’t
going to help
here.
Decide which
variable to
eliminate
–

Let’s do x
Now add together
4x - 2y = 5
-2x + y = 1
x2
-4x + 2y = 2
4x - 2y = 5
0+0=7
Not True
No Solutions