The Substitution Method A method to solve a system of linear equations in 2 variables.

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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

Choose either equation and solve for either variable.

(Choose the easiest one the solve. )

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

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

-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

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 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

y = 2x - 1

y = 2(19) - 1

x = 38-1=37

(x , y ) = (19,37)

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

q = 35 - d

q = 35 - 22

q = 13

# of dimes = 22# of quarters =13

the other equation10d +875-25d = 545

-15d + 875 = 545 -15d = -330

d = -330/-15 = 22

and solve.

The Substitution Method A method to solve a system of linear equations in 2 variables.

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