Question in description

Question

Jennifer and Cody are selling fruit for a school fundraiser. Customers can buy small boxes of oranges and large boxes of oranges. Jennifer sold 2 small boxes of oranges and 6 large boxes of oranges for a total of $118. Cody sold 7 small boxes of oranges and 12 large boxes of oranges for a total of $278. What is the cost of each of one small box of oranges and one large box of oranges?

Answer 1

X= small box

y= large box

 

Jennifers equation: 2x+6y=118

Codys equation: 7x+12y=278

 

There are 2 methods to solve this

 

Method 1:

Solve one equation for X

2x+6y=118

2x=118-6y

x=59-3y

 

Insert this equation in for x in Codys equation

 

7(59-3y)+12y=278

413-21y+12y=278

-9y=-135

y=15

 

insert 15 in for y

 

2x+6(15)=118

2x+90=118

2x=28

x=14

 

So small boxes are $14 and large boxes are $15

 

Method 2:

 

(2x+6y=118)*-2

7x+12y=278

 

   -4x-12y=-236

+  7x+12y=278

___________

    3x        = 42

              x=14

 

Then substitute in for x

 

2(14)+6y=118

28+6y=118

6y=90

y=15

 

So small boxes are $14 and large are $15

 

Answer 2

let s = small box and L = large box

Here's what we know:

2s + 6L = 118

7s + 12L = 278

 

We can solve this either with isolation or with elimination.

In isolation, we divide the first equation by 2 and move the L variable to the other side, isolating the s variable.

In elimination, we multiply the first equation by -2 and add/subtract the two equations together, thereby removing the L-variable

 

In this case, I'll go with elimination:

 

-2(2s + 6L = 118)

becomes:

-4s - 12L = -236

 7s + 12L = 278

------------------

3s = 42

Divide both sides by 3:

s = 14

 

2 * 14 + 6L = 118

28 + 6L = 118

6L = 90

L = 15

 

Check:

7 * 14 + 12 * 15 = 278

98 + 180 = 278

278 = 278