Find practice problems and solutions for these concepts at Solving Systems of Equations Algebraically Practice Problems.
You know how to solve systems of equations graphically. In this lesson, you will learn how to solve systems of equations algebraically. You will learn how to use the elimination method and the substitution method for solving systems of equations.
How to Use the Elimination Method
Graphs serve many useful purposes, but using algebra to solve a system of equations can be faster and more accurate than graphing a system. A system of equations contains equations with more than one variable. If you have more than one variable, you need more than one equation to solve for the variables. When you use the elimination method of solving equations, the strategy is to eliminate all the variables except one. When you have only one variable left in the equation, then you can solve it.
Example:
x + y = 10
x –y = 4
Add the equations. 
(x + y) + (x –y) = 10 + 4 

2x – 0y = 14 
Drop the 0y. 
2x = 14 
Divide both sides of the equation by 2. 

Simplify both sides of the equation. 
x = 7 
You have solved for the variable x. To solve for the variable y, substitute the value of the x variable into one of the original equations. It does not matter which equation you use.

x + y = 10 
Substitute 7 in place of the x variable. 
7 + y= 10 
Subtract 7 from both sides of the equation. 
7 – 7 + y = 10 –7 
Simplify both sides of the equation. 
y = 3 
You solve a system of equations by finding the value of all the variables. In the previous example, you found that x = 7 and y = 3. Write your answer as the ordered pair (7,3). To check your solution, substitute the values for x and y into both equations.
Check: x + y= 10
Substitute the values of the variables into the first equation. 
7 + 3 = 10 
Simplify. 
10 = 10 

x –y = 4 
Substitute the values of the variables into the second equation. 
7 – 3 = 4 
Simplify. 
4 = 4 
Did you get the right answer? Because you got true statements when you substituted the value of the variables into both equations, you solved the system of equations correctly. Try another example.
Example:
x + y = 6
–x + y = –4
Add the two equations. 
0x + 2y = 2 
Drop the 0x. 
2y = 2 
Divide both sides of the equation by 2. 

Simplify both sides of the equation. 
y = 1 
Use one of the original equations to solve for x. 
x + y = 6 
Substitute 1 in place of y. 
x + 1 = 6 
Subtract 1 from both sides of the equation. 
x + 1 – 1 = 6 – 1 
Simplify both sides of the equation. 
x = 5 
Write the solution of the system as an ordered pair: (5,1).
Check: x + y = 6
Substitute the values of x and y into the first equation. 
5 + 1 = 6 
Simplify. 
6 = 6 

–x + y = –4 
Substitute the values of x and y into the second equation. 
–(5) + 1 = –4 
Simplify. 
–4 = –4 
Did you get the right answer? Yes! You got true statements when you substituted the value of the variables into both equations, so you can be confident that you solved the system of equations correctly.
When You Can't Easily Eliminate a Variable
Sometimes, you can't easily eliminate a variable. Take a look at the following example. What should you do?
Example:
x + y = 24
2x + y = 3
If you were to add this system of equations the way it is, you would be unable to eliminate a variable. However, if one of the y variables were negative, you would be able to eliminate the y variable. You can change the equation to eliminate the y variable by multiplying one of the equations by –1. You can use either equation. Save time by choosing the equation that looks easier to manipulate.
Multiply both sides of the first equation by –1. 
–1(x + y) = –1 · 24 
Simplify both sides of the equation. 
–x –y = –24 
Add the second equation to the modified first equation. 
2x + y = 3 

x = –21 

x + y = 24 
Substitute the value of x into one of the original equations. 
–21 + y = 24 
Add 21 to both sides of the equation. 
–21 + 21 + y = 24 + 21 
Simplify both sides of the equation. 
y = 45 
The solution to the system of equations is (–21,45).
Example:
2x + y = 4
3x + 2y = 6
How will you eliminate a variable in this system? If you multiply the first equation by –2, you can eliminate the y variable. 
–2(2x + y) = –2 · 4 

–4x – 2y = –8 
Add the second equation to the modified first equation. 
3x + 2y = 6 

–x = –2 
Multiply both sides of the equation by –1. 
–1 · –x = –1 · –2 

x = 2 
Substitute 2 in place of the x variable in one of the original equations. 

2x + y = 4 

2 · 2 + y= 4 
Simplify. 
4 + y = 4 
Subtract 4 from both sides of the equation. 
4 – 4 + y = 4 – 4 
Simplify. 
y = 0 
The solution of the system is (2,0).
Example:
5x – 2y = 8
3x – 5y = 1
Are you ready for a challenge? To eliminate one of the variables in this system of equations, you will need to alter both equations. If you multiply the first equation by 5 and the second equation by –2, then you will be able to eliminate the y variable. Or, you could multiply the first equation by 3 and the second equation by –5 to eliminate the x variable. To decide which variable to eliminate, choose the one that looks like it will be the easier one to do.
Multiply the first equation by 5. 
5(5x – 2y) = 8 · 5 
Multiply the second equation by –2. 
–2(3x – 5y) = 1 · –2 
Simplify the first equation. 
25x – 10y = 40 
Simplify the second equation. 
–6x + 10y = –2 


Add both equations. 
19x = 38 
Divide both sides of the equation by 19. 

Simplify both sides of the equation. 
x = 2 
Substitute the 2 in place of the x variable into one of the original equations. 
5x – 2y = 8 

5 · 2 – 2y = 8 
Simplify. 
10 – 2y = 8 
Subtract 10 from both sides of the equation. 
10 – 10 – 2y = 8 – 10 
Simplify both sides of the equation. 
–2y = –2 
Divide both sides of the equation by –2. 

Simplify both sides of the equation. 
y = 1 
The solution of the system is (2,1).
Rearranging Equations
Sometimes, it is necessary to rearrange an equation before you can solve it using the elimination method. Take a look at this example.
Example:
2x = 2y + 6
3x = –3y + 3
Rearrange the first equation. 
2x = 2y + 6 
Subtract 2y from both sides of the equation. 
2x – 2y = 2y – 2y + 6 
Simplify. 
2x – 2y = 6 
Rearrange the second equation. 
3x = –3y + 3 
Add 3y to both sides of the equation. 
3x + 3y = –3y + 3y +3 
Simplify. 
3x + 3y = 3 
Your system has been altered to these equations. 
2x – 2y = 6 

3x + 3y = 3 
Multiply the first equation by 3. 
3(2x – 2y) = 3 · 6 
Multiply the second equation by 2. 
2(3x + 3y ) = 2 · 3 
Simplify both equations. 
6x – 6y = 18 

6x + 6y = 6 


Add both equations. 
12x = 24 
Divide both sides of the equation by 12. 

Simplify both sides. 
x = 2 
Substitute 2 in place of x in one of the original equations. 
2x = 2y + 6 

2 · 2 = 2y + 6 
Simplify. 
4 = 2y + 6 
Subtract 6 from both sides of the equation. 
4 – 6 = 2y + 6 – 6 
Simplify both sides of the equation. 
–2 = 2y 
Divide both sides by 2. 

Simplify. 
–1 = y 
The solution of the system is (2,–1).
How to Use the Substitution Method
Some equations are easier to solve using the substitution method instead of the elimination method. Study the following examples.
Example:
y = 2x
2x + y = 12
The first equation says that y = 2x. If you substitute 2x in place of y in the second equation, you can eliminate a variable. 
2x + 2x = 12 
Combine similar terms. 
4x = 12 
Divide both sides of the equation by 4. 

Simplify both sides of the equation. 
x = 3 
Substitute 3 in place of x in one of the original equations. 
y = 2x 

y = 2 · 3 

y = 6 
The solution of the system is (3,6).
Example:
x = 2y + 1
x + 3y = 16
Substitute 2y + 1 in place of x in the second equation. 
2y + 1 + 3y = 16 
Simplify. 
5y + 1 = 16 
Subtract 1 from both sides of the equation. 
5y + 1 – 1 = 16 – 1 
Simplify both sides of the equation. 
5y = 15 
Divide both sides of the equation by 5. 

Simplify both sides of the equation. 
y = 3 
Substitute 3 in place of y in one of the original equations. 
x = 2y + 1 

x = 2 · 3 + 1 

x = 7 
The solution for the system is (7,3).
Applications
Use a system of equations to solve the following word problems. First, work through the example problem given.
Example: Your book club is having a pizza party. The pepperoni pizza is $8, and the combination pizza is $12. You need 9 pizzas, and you have $80 to spend. How many of each kind can you get?
Let x = number of pepperoni pizzas.
Let y = number of combination pizzas.
Multiply the first equation by –8. 
–8(x + y) = –8 · 9 
Simplify. 
–8x – 8y = –72 
Add the second equation to the altered first equation. 
8x + 12y = 80 

4y = 8 
Divide both sides of the equation by 4. 

Simplify both sides of the equation. 
y = 2 
Substitute 2 in place of y in one of the original equations. 
x + y = 9 

x + 2 = 9 
Subtract 2 from both sides of the equation. 
x+ 2 – 2 = 9 – 2 
Simplify. 
x = 7 
You can get 7 pepperoni pizzas and 2 combination pizzas.
Skill Building until Next Time
What kind of problems in your everyday life could be solved using a system of equations? Use the word problems under the Applications heading to help you generate ideas. After you come up with an idea for a system of equations, go ahead and solve it using the elimination or substitution method. Then, check your answer as shown in this lesson to be sure you got the correct answer.

Find practice problems and solutions for these concepts at Solving Systems of Equations Algebraically Practice Problems.
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