Substitution method
System of linear equations can also be solved using the substitution method.We will show with examples.
Before you learn this lesson, make sure you understand how to solve
linear equations
Example #1: Solve the following system using the substitution method
x + y = 20
x − y = 10
Step 1
You have two equations. Pick either the first or the second equation and solve for either x or y
Since I am the one solving it, I have decided to choose the equation at the bottom (x − y = 10) and I will solve for x
x − y = 10
Add y to both sides
x − y + y = 10 + y
x = 10 + y
Step 2
Since you used the equation at the bottom to solve for x, you will substitute x into the equation on top (x + y = 20)
Using x + y = 20, erase x and write 10 + y since x = 10 + y
We get 10 + y + y = 20
10 + 2y = 20
Minus 10 from both sides
10 − 10 + 2y = 20 − 10
2y = 10
Divide both sides by 2
y = 5
Now you have y, you can replace its value into either equation to get x
Replacing y into x + y = 20 gives
x + 5 = 20
Minus 5 from both sides
x + 5 − 5 = 20 − 5
x = 15
The solution to the system is x = 15 and y = 5
Indeed 15 + 5 = 20 and 15 − 5 = 10
Example #2: Solve the following system using the substitution method
3x + y = 10
4x − 2y = 2
Step 1
You have two equations. Pick either the first or the second equation and solve for either x or y
I have decided to choose the equation on top (3x + y = 10) and I will solve for y
3x + y = 10
Subtract 3x from both sides
3x − 3x + y = 10 − 3x
y = 10 − 3x
Step 2
Since you used the equation on top to solve for y, you will substitute y into the equation at the bottom (4x − 2y = 2)
Using 4x − 2y = 2, erase y and write 10 − 3x keeping in mind that there is a multiplication between 2 and y
We get 4x − 2 ×(10 − 3x ) = 2
4x − 2 ×(10 − 3x ) = 2
4x − 20 + 6x = 2 (After multiplying 2 by 10 and 2 by 3x)
2x − 20 = 2
Add 20 to both sides
2x − 20 + 20 = 2 + 20
2x = 22
Divide both sides by 2
x = 11
Now you have x, you can replace its value into either equation to get y
Replacing x into 3x + y = 10 gives
3 × 11 + y = 10
33 + y = 10
Minus 33 from both sides
33 − 33 + y = 10 − 33
y = 23
The solution to the system is x = 11 and y = 23
Indeed,
3 × 11 + 23 = 33 + 23 = 10 and 4 × 11 − 2 × 23 = 44 + 46 = 2
Y
ou should have noticed that the reason we call this method the substitution method is because after you have solve for a variable in one equation,
you substitute the value of that variable into the other equation

Nov 15, 18 05:01 PM
Modeling multiplication with number counters  Learning multiplication is fun!
Read More
New math lessons
Your email is safe with us. We will only use it to inform you about new math lessons.