Q. 140 - Intermediate Algebra

Question

In the following exercise, translate to a system of equations and solve.   

Hannah has to make twenty-five gallons of punch for a potluck. The punch is made of soda and fruit drink. The cost of the soda is \($1.79$. per gallon and the cost of the fruit drink is \)$2.49$ per gallon. Hannah’s budget requires that the punch cost $$2.21$ per gallon. How many gallons of soda and how many gallons of fruit drink does she need?

Step-by-Step Solution

Answer

Hannah needs $10$ gallons of soda and $15$ gallons of fruit mix.

Step 1. Identify and name what we are looking for

We need to find the amount of soda and fruit mixed required by Hannah.

Let Hannah needs $x$ gallons of soda and $y$ gallons of fruit mix.

Step 2. Form the equation

The punch is of $25$ gallons, so soda and fruit mix combined is equal to $25$, so

$x+y=25 ...\left(1\right)$

The cost of soda is $$1.79$ per gallon and the cost of fruit mix is $$2.49$ per gallon, while the cost of punch is $$2.21$ per gallon. So the second equation can be written as

$$\begin{gathered} 1.79x+2.49y=2.21\times 25 \\ 1.79x+2.49y=55.25 ...\left(2\right) \end{gathered}$$

Step 3. Solve using substitution

Solve the first equation for $y$

$\begin{array}{rcl}x+y& =& 25\\ x+y-x& =& 25-x\\ y& =& 25-x ...\left(3\right)\end{array}$

Now using the third equation substitute $25-x$ for $y$ in the second equation and solve for $x$

$\begin{array}{rcl}1.79x+2.49y& =& 55.25\\ 1.79x+2.49(25-x)& =& 55.25\\ 1.79x+62.25-2.49x& =& 55.25\\ 1.79x-2.49x+62.25-62.25& =& 55.25-62.25\\ -0.7x& =& -7\\ x& =& 10\end{array}$

Step 4. Find the value of y

Substitute $10$ for $x$ in the third equation

$$\begin{gathered} y=25-x \\ y=25-10 \\ y=15 \end{gathered}$$

Thus the punch contains $10$ gallons of soda and $15$ gallons of fruit mix.

Step 5. Check the solution

Substitute $10$ for $x$ and $15$ for $y$ in the first equation formed.

$\begin{array}{rcl}x+y& =& 25\\ 10+15& =& 25\\ 25& =& 25\end{array}$

It is a true statement.

Again, substitute the values in the second equation formed.

$\begin{array}{rcl}1.79x+2.49y& =& 55.25\\ 1.79·10+2.49·15& =& 55.25\\ 17.9+37.35& =& 55.25\\ 55.25& =& 55.25\end{array}$

This is also a true statement.

So the point satisfies both the equations.