This video shows how linear systems can be applied to word problems. After you finish this lesson, view all of our Algebra 1 lessons and practice problems.

When you encounter a word problem, it is important to write down all the important information given to you.
You would then use that information to write out the equations needed to solve the problem.
In this case, its solving by elimination, where you solve for one variable by eliminating the other variable, and then plugging your answer back into an original equation to find the other variable.
Be sure to write in the units used by the word problems.

Example:

Li and Jordan went to the store. Li bought 3 boxes of cereal and 5 muffins for $22. Jordan bought one box of cereal and eight muffins for $20.
What is the cost of a box of cereal?
What is the cost of a muffin?

 3c + 5m = 22
 \underline{-3( c + 8m = 20 )}

 3c + 5m = 22
 \underline{-3c - 24m = -60}
 -19m = -38
 m = 2
 \downarrow
 c + 8m = 20
 c + 8(2) = 20
 c + 16 = 20
 c = 4

Cost of a box of cereal = $4
Cost of a muffin = $2

Applications of Linear Systems

Video-Lesson Transcript

Let’s go over application of linear systems.

At this point, we’ll solve the problem using substitution, elimination or graphing.

For example:

Li and Jordan went to the store. Li bought 3 boxes of cereal and 5 boxes of muffins for \$22. Jordan bought one box of cereal and eight muffins for \$20.
What is the cost of a box of cereal?
What is the cost of the muffin?

First, let’s define our variables.

cost of a box of cereal = c
cost of muffin = m

Let’s equate what Li bought.

3c + 5m = 22

Let’s go to Jordan.

c + 8m = 20

Now we have our system equations.

Let’s solve this by elimination first.

Let’s multiply the second equation by -3 so we can cancel the cost of cereal.

3c + 5m = 22
-3 (c + 8m) = -3 (20)
3c + 5m = 22
-3c - 24m = -60

Now let’s add the two

-19m = -38

Solve for m by dividing both sides by -19.

\dfrac{-19m}{-19} = dfrac{-38}{-19}
m = 2

So the cost of muffin is \$2.

Let’s substitute this value to find the value of cereal.

Let’s use the second equation to do this.

c + 8m = 20
c + 8(2) = 20
c + 16 = 20

Let’s solve for c

c + 16 - 16 = 20 - 16
c = 4

So the cost of cereal is \$4.