Like?

# How Do I Use a System of Equations?

Start Your Free Trial To Continue Watching
As a member, you'll also get unlimited access to over 8,500 lessons in math, English, science, history, and more. Plus, get practice tests, quizzes, and personalized coaching to help you succeed.
Free 5-day trial
It only takes a minute. You can cancel at any time.
Already registered? Login here for access.
Start your free trial to take this quiz
As a premium member, you can take this quiz and also access over 8,500 fun and engaging lessons in math, English, science, history, and more. Get access today with a FREE trial!
Free 5-day trial
It only takes a minute to get started. You can cancel at any time.
Already registered? Login here for access.
1. 0:05 Building a System of Equations
2. 4:12 Solving with Substitution
3. 6:17 Solving with Elimination
4. 8:16 Lesson Summary
Share
Show Timeline

### Jeff Calareso

Jeff has taught high school English, math and other subjects. He has a master's degree in writing and literature.

There are a few classic algebra word problems, such as the one about two trains traveling at different speeds. In this lesson, you'll learn how to take a word problem and convert it into the system of equations that will allow you to find the answer using either substitution or elimination.

## Building a System of Equations

Probably the best known algebra word problem (which often seems to have been a traumatic experience for people) has to do with two trains leaving the station going opposite directions. It usually asks something along the lines of how fast they were going or how far apart they were. But I like to make math as un-traumatic as possible, so we're going to stay away from any train problems and instead, we're going to look at a difference classic word problem.

Let's say you're hosting a Super Bowl party and have been given \$68 by all your friends that are coming to spend on pizza. You're going to get either cheese pizza or pepperoni pizza. Cheese pizzas cost \$8 each, and the pepperoni pizzas cost \$11 each. Since you might as well spend all the money (it's your friends', not yours) and you've also decided that 7 pizzas is the perfect amount of food, the question is how many of the seven pizzas should be cheese, and how many of them should be pepperoni?

We could try to guess and check our way through this problem, which might work for this one, but that's probably not going to work for most problems like this. So we'd rather use this example as a chance to learn how to do it with math. But in order to do it with math, we have to first change the words into mathematical equations. This can oftentimes be the hardest part.

In order to set up those equations, we need to figure out what the variables we're looking for are. In order to that, we're going to have to go back to the problem and figure out what it's really asking us for. This problem is asking us for how many cheese pizzas and how many pepperoni pizzas should we buy. Which means our variables should be the number of cheese pizzas (C) and the number of pepperoni pizzas (P). When we solve for those two things, we'll know how many we should buy.

So now that we have our variables, we have to come up with equations that represent the situation. This problem is going to need two equations to solve it because there are two variables, which means that this is a system of equations, because there's going to be more than one.

The easier equation to come up with uses the fact that we know we have to buy a total of 7 pizzas. We've decided that 7 pizzas is the perfect amount. Since C is the number of cheese and P is the number of pepperoni, and we know that the total has to be 7, my first equation is the number of cheese (C) plus the number of pepperoni (P) must equal 7: C + P = 7.

The harder equation to come up with has to do with the price and how much money we have and how much money different pizzas cost. What I want to do is set up an equation that equals 68. It should say that we are going to spend \$68. This means I have to figure out how much money I'm spending on my cheese pizzas and how much money I'm spending on my pepperoni pizzas.

Well, if we just think about this logically, if I bought one cheese pizza, it would cost me \$8. And if I bought two, it would cost me \$16. If I bought three, it would cost me \$24, and so on. This means that the total money I spend on my cheese pizzas is simply 8 times the number of cheese pizzas I buy (8C). A very similar expression could be derived for the amount of money you spend on pepperoni pizzas by doing 11 times the number of pepperoni pizzas you buy (11P).

So 8C represents the money you spend on cheese pizzas and 11P represents the money you spend on pepperoni pizzas. If you add those two amounts together, that's the total amount we spend. And that has to be \$68. So our second equation, the little more complicated one, is 8C + 11P = \$68.

## Solving a System of Equations with Substitution

Now that we have the system set up, we can actually begin to solve it. We're going to go over 2 ways of solving a system of equations, the first of which is called substitution. It is called this because we're going to take one expression from one equation and substitute it in for one of the variables in our second equation.

For example, we can take the equation that says C + P = 7, telling us the total number of pizzas we're getting, and we can solve this equation for P by undoing C from both sides with subtraction. We find an equivalent equation that says that P = 7 - C.

Now that I know what P equals, and since I want these two equations to be the same, I can substitute into the second equation what I now know what P is. So instead of writing 8C + 11P = \$68, I can write 8C + 11(7 - C) = \$68. Now I only have one variable in one equation, which is solvable for that variable.

I can use inverse operations to get the C by itself. First, you distribute the 11 to both terms: 8C + 77 - 11C = \$68. Then, you can combine like terms, grouping the Cs together because they're on the same side of the equals sign: -3C + 77 = \$68. Next, you undo an addition of 77 with a subtraction of 77: -3C = -9. Last but not least, you undo a multiplication of -3 with a division of -3 to both sides. You find that C, the number of cheese pizzas we're going to get, is 3: C = 3.

Now that I know the number of cheese pizzas, it's a pretty easy process to just take that and substitute back into the first equation: P = 7 - C. I now know that C is 3, which means that P = 7 - 3. This means that P is 4. I now know how many pepperoni pizzas we should buy.

## Solving a System of Equations with Elimination

So that's the substitution method to solving a system of equations. But there's another method called the elimination method, which is really nice to use when both of my equations are in standard form. This means that both of my equations have the C's and the P's on the same side.

## Unlock Content Over 8,500 lessons in all major subjects

Get FREE access for 5 days,
just create an account.

No obligation, cancel anytime.

Want to learn more?

Select a subject to preview related courses:

###### People are saying…

"This just saved me about \$2,000 and 1 year of my life." — Student

"I learned in 20 minutes what it took 3 months to learn in class." — Student

See more testimonials

Yes No

### What didn't you like?

What didn't you like?

Congratulations! You've reached the last video in the chapter.
###### Create your Account

Sign up now for your account. Get unlimited access to 8,500 lessons in math, English, science, history, and more.

###### Meet Our Instructors

Meet all 53 of our instructors

###### Related Articles
• Related
• Recently Updated
• Popular