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## Solving Word Problems

To solve a word problem using a system of equations, it is important to;

– Identify what we don’t know

– Declare variables

– Use sentences to create equations

An example on how to do this:

Mary and Jose each bought plants from the same store. Mary spent $188 on 7 cherry trees and 11 rose bushes. Jose spent $236 on 13 cherry trees and 11 rose bushes. Find the cost of one cherry tree and the cost of one rose bush.

Cost of a cherry tree:

Cost of a rose bush:

7 cherry trees and 11 rose bushes = $188

The y-values cancel each other out, so now you are left with only x-values and real numbers.

Then, you plug in your x-value into an original equation in order to find the y-value.

Cost of a cherry tree: $8

Cost of a rose bush: $12

## Examples of Word Problems – System Of Equations

### Example 1

Three coffees and a cupcake cost a total of dollars. Two coffees and four cupcake cost a total of dollars. What is the individual price for a single coffee and a single cupcake?

Let’s solve this by following steps.

1. What we don’t know:

Cost of a single coffee

Cost of single cupcake

2. Declare variables:

Cost of a single coffee=

Cost of single cupcake=

3. Use sentences to create equations.

Three coffees and a cupcake cost a total of dollars.

Two coffees and four cupcake cost a total of dollars.

Now, we have a system of equations:

Let’s solve for one of the variables in one of the equations and then use that to substitute into the other.

Now, solve for the value of using the first equation.

Let’s solve the value of by substituting the value of to the bottom equation.

Distribute to each terms inside the parenthesis

Combine like terms

Now, let’s isolate the by subtracting on both sides.

Then divide both sides by ,

And we’ll have

Then, let’s plug the value of into one equation to get the value of .

Cost of a single coffee=

Cost of single cupcake=

### Example 2

The senior class at High School A rented and filled vans and buses with students. High School B rented and filled vans and bus with students. Every van had the same number of students in it as did the buses. Find the number of students in each van and in each bus.

Let’s solve this by following steps.

1. What we don’t know:

Students in each van

Students in each bus

2. Declare variables:

Students in each van=

Students in each bus=

3. Use sentences to create equations.

High School A rented and filled 8 vans and 8 buses with 240 students.

High School B rented and filled 4 vans and 1 bus with 54 students.

Now, we have a system of equations:

Let’s solve for one of the variables in one of the equations and then use that to substitute into the other.

Now, solve for the value of using the second equation.

Let’s solve the value of by substituting the value of to the bottom equation.

Distribute to each terms inside the parenthesis

Combine like terms

Now, let’s isolate the by subtracting on both sides.

Then divide both sides by ,

And we’ll have

Then, let’s plug the value of into one equation to get the value of .

Students in each van=

Students in each bus=

## Video-Lesson Transcript

To solve a word problem using system of equations, it is important to:

1. Identify what we don’t know

2. Declare variables.

3. Use sentences to create equations.

Let’s have an example:

Mary and Jose each bought plants from the same store. Mary spent on cherry trees and rose bushes. Jose spent on cherry trees and rose bushes. Find the cost of one cherry tree and the cost of one rose bush.

Let’s solve this by following steps above.

1. What we don’t know:

cost of a cherry tree

cost of a rose bush

2. Declare variables:

cost of a cherry tree :

cost of a rose bush :

3. Use sentences to create equations.

For Mary:

cherry trees and rose bushes

For Jose:

cherry trees and rose bushes

Now, we have a system of equations

We can solve this by process of substitution, elimination or fraction.

Since the value of is the same for both equations, let’s do the process of elimination.

First, let’s multiply the first equation by

Here we’ll have a negated equation

Let’s do the process of elimination now

We’ll have

Then, let’s isolate by dividing both sides by

Now, we have

Remember, our declared variable?

cost of a cherry tree :

Since

Now we can say that

cost of a cherry tree :

Now, let’s solve for the value of by getting one equation and plugging the value of .

Let’s use the first equation to plug in

Let’s isolate by subtracting on both sides of the equation

Then divide by

And we get

Now, we know that cost of a rose bush is .