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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

2. Declare variables:

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

Word Problems – System of Equations

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.