Transcript Document

The Substitution
Method
A method to solve a
system of linear
equations in 2 variables
When you “Solve a
system of equations”
you are looking for a
solution that will
solve every equation
in the system (group).
???
A linear equation
Ax + By = C
has an infinite number
of solution?
???
How do we start with
two equations , each
having an infinite
number solutions, and
find the common
solution (if any)???
This Method will
create one combined
equation with only one
variable. This is the
kind of equation that
you can solve!
Substitution Method - Step One
Solve one equation for
one of the variables
Choose either equation and
solve for either variable.
(Choose the easiest one the solve. )
Step One - Solve one equation
for one of the variables
1
2x + y = 53
2
x + 5y = 139
Choose either equation and solve for either variable.
(Choose the easiest one the solve. )
Step One - Solve one equation for one of the variables
In this
problem
you could
have
solved
equation
#1 for y
or solved
equation
# 2 for x.
1
-2x
-2x
2x + y = 53
y = -2x +53
-5y -5y
2
x + 5y = 139
x = - 5y +139
Step One - Solve one equation
for one of the variables
Substitution Method - Step Two
Substitute this
expression in the other
equation and solve.
Substitute this expression in the other equation and solve.
If you solve for y in the first equation take this expression and
substitute it in for y in the 2nd equation
1
2
joined
2x + y = 53
y = -2x +53
x + 5y = 139
x + 5 (-2x +53) = 139
This will create one
combined equation
with only one variable.
This is the kind of
equation that you can
solve!
Substitute this expression in the other equation and solve.
Now solve for x
1
2
joined
2x + y = 53
x + 5y = 139
y = -2x +53
x + 5 (-2x +53) = 139
x + -10x+265 = 139
-9x + 265 = 139
-9x = -126
x = -126/-9=14
After solving the combined equation ….
Find the
corresponding value
of the other variable.
(Substitute the value you found in
step 2 to back into the equation)
Substitute the value you
found for the first
variable back into one of
the original equations
1
2
2x + y = 53 y = -2x +53
x + 5y = 139
x + 5 (-2x +53) = 139
x + -10x+265 = 139
-9x + 265 = 139
-9x = -126
x = -126/-9 = 14
y = -2x +53
y = -2(14)+53
y = -28+53=25
From the last step you
found that x was 14.
Take this value and
plug it back into one of
the original equations
and find y.
(x , y ) = (14,25)
Solve one of the
equations for one
of the variables
1
2
x = 4y +5
x - 4y = 5
3x + 2y = 113
3 (4y +5) + 2y = 113
12y +15+2y = 113
14y + 15 = 113
and solve.
14y = 98
y = 98/14 = 7
(x , y ) = (33,7)
Substitute this
expression into the
other equation
x = 4y +5
x = 4(7) +5
x = 28+5=33
Substitute the value you
found for the first
variable back into one of
the original equations
Step One - Solve one equation
for one of the variables
Step Two - Substitute this
expression in the other
equation and solve.
Step Three - Find the other
variable (Substitute value back
into one of the equations)
I’m thinking of two
numbers. One number
is one less then twice
the other. The
difference of the
numbers is 18. Find the
numbers.
I’m thinking of two numbers. One
number is one less then twice the
other. The difference of the
numbers is 18. Find the numbers.
Let x and y represent the numbers.
Write two equations to represent
the relationships.
y=2x-1
y-x=18
Solve one of the
equations for one
of the variables
1
2
y = 2x-1
y - x = 18
2x - 1 - x = 18
x-1 = 18
x = 19
and solve.
(x , y ) = (19,37)
y = 2x - 1
Substitute this
expression into the
other equation
y = 2x - 1
y = 2(19) - 1
x = 38-1=37
Substitute the value you
found for the first
variable back into one of
the original equations
John had all dimes and
quarters worth $5.45
If he had 35 coins in all,
find out how many of
each coin he had.
John had all dimes and quarters
worth $5.45 If he had 35 coins in
all, find out how many of each coin
he had.
Let d = the # of dimes and
q = the # of quarters
Write two equations.
d+q=35 10d+25q=545
Solve one of the
equations for one
of the variables
1
2
d + q = 35 q = 35 - d
10d + 25q = 545
10d + 25(35 - d) = 545
10d +875-25d = 545
-15d + 875 = 545
and solve.
-15d = -330
d = -330/-15 = 22
# of dimes = 22
# of quarters =13
Substitute this
expression into the
other equation
q = 35 - d
q = 35 - 22
q = 13
Substitute the value you
found for the first
variable back into one of
the original equations