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The Substitution
MethodA 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 = Chas 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
2x + y = 53
x + 5y = 139
1
2
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.
-5y -5y
-2x -2x
x = - 5y +139
y = -2x +532x + y = 531
x + 5y = 1392
Substitution Method - Step Two
Step One - Solve one equation for one of the variables
Substitute this expression in the other equation and solve.
Substitute this expression in the other equation and solve.
2x + y = 531
x + 5y = 1392
y = -2x +53
If you solve for y in the first equation take this expression and substitute it in for y in the 2nd equation
x + 5 (-2x +53) = 139joined
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.
2x + y = 531
x + 5y = 139y = -2x +53Now solve for x
x + 5 (-2x +53) = 139joined
x + -10x+265 = 139-9x + 265 = 139
2
-9x = -126x = -126/-9=14
Find the corresponding value of the other variable.
After solving the combined equation ….
(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
y = -2x +532x + y = 531
x + 5y = 1392
x + 5 (-2x +53) = 139x + -10x+265 = 139
-9x + 265 = 139-9x = -126
x = -126/-9 = 14
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.
y = -2x +53y = -2(14)+53
y = -28+53=25
(x , y ) = (14,25)
x - 4y = 51
3x + 2y = 113x = 4y +5
3 (4y +5) + 2y = 113
2
Substitute the value you found for the first variable back into one
of the original equations
x = 4y +5
x = 4(7) +5
x = 28+5=33
(x , y ) = (33,7)
Solve one of the equations for one of the variables
Substitute this expression into
the other equation12y +15+2y = 113
14y + 15 = 113 14y = 98
y = 98/14 = 7
and solve.
3 ( ) + 2y = 113
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
y = 2x-11
y - x = 18y = 2x - 1
2x - 1 - x = 18
2
Substitute the value you found for the first variable back into one
of the original equations
y = 2x - 1
y = 2(19) - 1
x = 38-1=37
(x , y ) = (19,37)
Solve one of the equations for one of the variables
Substitute this expression into
the other equationx-1 = 18
x = 19and solve.
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
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.
d + q = 351
10d + 25q = 545q = 35 - d
10d + 25(35 - d) = 545
2
Substitute the value you found for the first variable back into one
of the original equations
q = 35 - d
q = 35 - 22
q = 13
# of dimes = 22# of quarters =13
Solve one of the equations for one of the variables
Substitute this expression into
the other equation10d +875-25d = 545
-15d + 875 = 545 -15d = -330
d = -330/-15 = 22
and solve.
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