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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: $x$
Cost of a rose bush: $y$
7 cherry trees and 11 rose bushes = \$188
$7x + 11y = 188$
$13x + 11y = 236$

$7x + 11y = 188$
$\underline {-13x - 11y = -236 }$

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

$-6x = -48$
$x = 8$

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

$7x + 11y = 188$
$7(8) + 11y = 188$
$56 + 11y = 188$
$11y = 132$
$y = 12$

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 $7$ dollars. Two coffees and four cupcake cost a total of $8$ 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

Cost of a single coffee= $x$

Cost of single cupcake= $y$

3. Use sentences to create equations.

Three coffees and a cupcake cost a total of $7$ dollars.

$3x+y=7$

Two coffees and four cupcake cost a total of $8$ dollars.

$2x+4y=8$

Now, we have a system of equations:

$3x+y=7$ $2x+4y=8$

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 $y$ using the first equation.

$3x+y=7$ $y=7-3x$

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

$2x+4(7-3x)=8$

Distribute $4$ to each terms inside the parenthesis

$2x+28-12x=8$

Combine like terms

$28-10x=8$

Now, let’s isolate the $y$ by subtracting $28$ on both sides.

$28-10x-28=8-28$ $-10x=-20$

Then divide both sides by $-10$,

$\dfrac{-10x}{-10} = \dfrac{-20}{-10}$

And we’ll have

$x = 2$

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

$3x+y=7$ $3(2)+y=7$ $6+y=7$ $y=1$

Cost of a single coffee= $\2$

Cost of single cupcake=$\1$

### Example 2

The senior class at 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. 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= $v$

Students in each bus= $b$

3. Use sentences to create equations.

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

$8v+8b=240$

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

$4v+b=54$

Now, we have a system of equations:

$8v+8b=240$ $4v+b=54$

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 $b$ using the second equation.

$4v+b=54$ $b=54-4v$

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

$8v+8(54-4v)=240$

Distribute $8$ to each terms inside the parenthesis

$8v+432-32v=240$

Combine like terms

$432-24v=240$

Now, let’s isolate the $v$ by subtracting $432$ on both sides.

$432-24v-432=240-432$ $-24v=-192$

Then divide both sides by $-24$,

$\dfrac{-24}{-24} = \dfrac{-192}{-24}$

And we’ll have

$v = 8$

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

$4v+b=54$ $4(8)+b=54$ $32+b=54$ $b=22$

Students in each van= $8$

Students in each bus= $22$

## 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 $\188$ 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.

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 : $x$
cost of a rose bush : $y$

3. Use sentences to create equations.

For Mary:
$7$ cherry trees and $11$ rose bushes $= \188$
$7x + 11 y = 188$

For Jose:
$13$ cherry trees and $11$ rose bushes $= \236$
$13x + 11 y = 236$

Now, we have a system of equations

$7x + 11y = 188$
$13x + 11y = 236$

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

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

First, let’s multiply the first equation by $-1$

$-1 (13x + 11y = 236)$

Here we’ll have a negated equation

$-13x - 11y = -236$

Let’s do the process of elimination now

$7x + 11y = 188$
$+ {-13x} + 11y = 236$

We’ll have

$6x = -48$

Then, let’s isolate $x$ by dividing both sides by $-6$

$\dfrac{-6x}{-6} = \dfrac{-48}{-6}$

Now, we have

$x = 8$

Remember, our declared variable?

cost of a cherry tree : $x$

Since $x = 8$

Now we can say that

cost of a cherry tree : $= 8$

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

Let’s use the first equation to plug in $x = 8$

$7x + 11y = 188$ $7(8) + 11y = 188$ $56 + 11y = 188$

Let’s isolate $y$ by subtracting $56$ on both sides of the equation

$56 - 56 + 11y = 188 - 56$ $11y = 132$

Then divide by $11$

$\dfrac{11y}{11} = \dfrac{132}{11}$

And we get

$y = 12$

Now, we know that cost of a rose bush is $12$.