Q 193 - Intermediate Algebra

Question

In the following exercises, solve the given problem.  

The church youth group is selling snacks to raise money to attend their convention. Amy sold $2$ pounds of candy, $3$ boxes of cookies and $1$ can of popcorn for a total sales of \($65$. Brian sold $4$ pounds of candy, $6$ boxes of cookies and $3$ cans of popcorn for a total sales of \)$140$. Paulina sold $8$ pounds of candy, $8$ boxes of cookies and $5$ cans of popcorn for a total sales of $$250$. What is the cost of each item? 

Step-by-Step Solution

Answer

The cost of one pound of candy is $$20$.

The cost of one box of cookies is $$5$.

The cost of one can of popcorn is $$10$.

Step 1. Identify and name what we are to find

We need to find the cost of candy, cookies, and popcorn.

Let $x$ represent the cost of one pound of candy, $y$ represents the cost of one box of cookies, and $z$ represents the cost of one can of popcorn.

Step 2. Form the equations

Amy sold $2$ pounds of candy, $3$ boxes of cookies and $1$ can of popcorn for a total sales of $$65$. So the first equation is

$2x+3y+z=65 ...\left(1\right)$

Brian sold $4$ pounds of candy, $6$ boxes of cookies and $3$ cans of popcorn for a total sales of $$140$. So the second equation is

$4x+6y+3z=140 ...\left(2\right)$

Paulina sold $8$ pounds of candy, $8$ boxes of cookies and $5$ can of popcorn for a total sales of $$250$. So the third equation is

$8x+8y+5z=250 ...\left(3\right)$ 

Step 3. Solve by elimination

Multiply the first equation with $-4$ and add it with the third equation

$\begin{array}{rcl}-4(2x+3y+z)+(8x+8y+5z)& =& -4\times 65+250\\ -8x-12y-4z+8x+8y+5z& =& -260+250\\ -4y+z& =& -10 ...\left(4\right)\end{array}$

Multiply the second equation by $-2$ and add it with the third equation

$\begin{array}{rcl}-2(4x+6y+3z)+(8x+8y+5z)& =& -2\times 140+250\\ -8x-12y-6z+8x+8y+5z& =& -280+250\\ -4y-z& =& -30 ...\left(5\right)\end{array}$

Step 4. Find the value of y

Add the fourth and fifth equation

$\begin{array}{rcl}-4y+z+(-4y-z)& =& -10+(-30)\\ -4y+z-4y-z& =& -10-30\\ -8y& =& -40\\ y& =& 5\end{array}$

So, the cost of one box of cookies is $$5$.

Step 5. Find the value of z

Substitute $5$ for $y$ in the fourth equation

$\begin{array}{rcl}-4y+z& =& -10\\ -4\times 5+z& =& -10\\ -20+z+20& =& -10+20\\ z& =& 10\end{array}$

So, the cost of one can of popcorn is $$10$.

Step 6. Find the value of x

Substitute $5$ for $y$ and $10$ for $z$ in the first equation

$\begin{array}{rcl}2x+3y+z& =& 65\\ 2x+3\times 5+10& =& 65\\ 2x+15+10& =& 65\\ 2x+25-25& =& 65-25\\ 2x& =& 40\\ x& =& 20\end{array}$

So, the cost of one pound of candy is $$20$.