Did you ever encounter a problem where you seem to have lots of information, but still have a couple of critical numbers missing? In this lesson, we’ll practice solving word problems that require us to set up systems of equations.

Systems of Equations

Lots of things have systems. For example, there’s a knitter’s system of organizing yarn by color or a chef’s system of chopping and organizing ingredients before cooking. Even when society breaks down, systems still matter. Let’s say it’s the zombie apocalypse. You’ll stand a better chance of surviving if you use systems.

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This includes not only systems of avoiding zombies. It also means systems of equations. A system of equations is a group of two or more equations with the same variables. There are many different kinds of word problems involving systems of equations. In this lesson, we’ll focus on a few of the most common types of these problems. Remember, though, that the principles at work are the same in all of them.

Counting Practice

Let’s start with one that could save your life. It’s the zombie apocalypse. You try to take shelter at Farmer Zed’s place, but his barn is full of zombie animals. You know he had 20 animals, a mix of chickens and pigs. Your recon scout tells you he counts 50 legs, but can’t see what they are. You hope there are more chickens, because zombie chickens are easier to kill. Zombie pigs? Not so much.

Okay, math can help. First, we need our variables. What don’t we know? The number of chickens and pigs. Let’s use c for chickens and p for pigs.

Next, we need our equations. We know c + p = 20. That’s all the chickens and pigs combined. We know the chickens should have 2 legs and the pigs should have 4, supposing none were gnawed off by zombie ducks. So, 2c + 4p = 50.

To solve, let’s rearrange this first equation to c = 20 – p, then substitute 20 – p for c in the second problem. This is the substitution method. So, that’s 2(20 – p) + 4p = 50. We get 40 – 2p + 4p = 50. That becomes 2p = 10. So, p = 5.

Let’s plug in 5 for p in c + p = 20. So, c = 15. Let’s check our math by plugging p = 5 and c = 15 into the second equation. That’s 2(15) + 4(5) = 50. 30 + 20 = 50. Okay, so 5 zombie pigs and 15 zombie chickens.

Money Practice

Let’s try a problem involving money. In this post-apocalyptic hellscape you call home, you need two things most of all: bullets and cookies. Look, it may be the end times, but you still have a sweet tooth. Fortunately, you encounter some delusional retail workers who are convinced the monetary system should still function. I guess the zombies aren’t much different than Black Friday shoppers to them.

Unfortunately, their cash registers no longer work, so you have no receipt. But, you know you spent $615.00 on a total of 135 $3.00 bullets and $8.00 cookies. How many of each did you get?

Let’s make our variables b for bullets and c for cookies. That’s good enough for me. What are our equations? You got 135 total items. So, b + c = 135. And, with a money problem, we can multiply the cost of each item times the number of the item to get its total cost. So, 3b is the cost of $3.00 bullets, and 8c is the cost of $8.00 cookies. You spent $615.00, so 3b + 8c = 615.

Let’s use substitution again. We can make b + c = 135 into b = 135 – c. Then, substitute into the equation to get 3(135 – c) + 8c = 615. That simplifies to 405 – 3c + 8c = 615. 8c – 3c is 5c. And, 615 – 405 is 210. So, 5c = 210. Divide by 5, and get c = 42.

Plug 42 in for c in b + c = 135, and b = 93. Let’s check for zombies. No zombies? Okay, let’s check our work in the second equation. 3b + 8c = 615. That’s 3(93) + 8(42) = 615. 3 * 93 is 279. 8 * 42 is 336. 279 + 336 is 615. So, we have 93 bullets and 42 cookies. Hmm, that might last a few days.

Percent Practice

Let’s try a percent problem. These can seem like tricky systems of equations questions, but they’re really just using the same principles. Let’s say you meet up with a chemist who has the key to stopping the zombification process. She needs a water-based solution with 14% vinegar. Yep, vinegar. Vinegar does all kinds of useful things.

Unfortunately, you only have vinegar in 8% and 24% solutions, and it only works at 14%. If you want 300 gallons of 14% solution, how much 8% and 24% will you need?

Let’s use x and y. x can be the number of gallons we’ll need of the 8% solution, and y is the same for the 24%. We know we want 300 gallons. So, x + y = 300. That’s our first equation.

We also know we want the amount of vinegar in the 8% solution plus the amount in the 24% solution to equal 14% in the 300 gallons. We can rewrite what I just said as .08x + .24y = .14 * 300. That’s the number of gallons of 8% times the 8%, which is .08, plus the number of y gallons times its concentration at 24%, equaling the final 300 gallons at 14 zombie-stopping percent.

Let’s make that first equation x = 300 – y, and substitute it into the second equation to get .08(300 – y) + .24y = 42. Simplify that to 24 – .08y + .24y = 42. Let’s get the ys together, and move the 24 over to get .16y = 18. Divide by .16 and get y = 112.5. Plug that into x + y = 300, and x = 187.5.

Again, check for zombies. Good? And cookies? Running a little low, but if we get this finished, more cookies await. Ok, let’s plug 187.5 and 112.5 into our second equation. That’s .08(187.5) + .24(112.5) = 42. .08 * 187.5 is 15. And, .24 * 112.5 is 27. 15 + 27? 42. Zombies? Vanquished!

Lesson Summary

In summary, a system of equations is simply a group of equations with the same variables. Problems involving systems of equations are useful whether you’re organizing your yarn collection or battling the zombie threat. Just identify your variables, set up your equations, then solve for your variables. And, watch out for zombie ducks – they like to gnaw!

Learning Outcome

You should be able to solve word problems that involve setting up a system of equations after watching this video lesson.