Question

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

Marissa wants to blend candy selling for \($1.80$per pound with candy costing \)$1.20$per pound to get a mixture that costs her $$1.40$per pound to make. She wants to make $90$pounds of the candy blend. How many pounds of each type of candy should she use?

Short answer

The number of pounds of candy selling is $30$and the number of pounds of candy costing is$60$.

Step-by-step solution

Step 1. Identify and name what we are looking for  

We need to find the pounds of each type of candy.

Let $x$represents the number of pounds of selling and $y$represents the number of pounds of costing.

Step 2. Form the equations

The total blend is of $90$pounds, so an equation can be written as

localid="1644509243582" $x+y=90...\left(1\right)$

Marissa blends candy selling for $$1.80$per pound with candy costing $$1.20$per pound to get a mixture that costs $$1.40$pounds, so an equation can be written as

$$\begin{gathered} 1.80x+1.20y=1.40\times 90 \\ 1.80x+1.20y=126...\left(2\right) \end{gathered}$$

Step 3. Solve using substitution

Solve the first equation for$y$

$\begin{array}{rcl}x+y& =& 90\\ x+y-x& =& 90-x\\ y& =& 90-x...\left(3\right)\end{array}$

Using the third equation, substitute $90-x$for $y$in the second equation and solve for $x$

$\begin{array}{rcl}1.80x+1.20y& =& 126\\ 1.80x+1.20(90-x)& =& 126\\ 1.80x+108-1.20x& =& 126\\ 1.80x-1.20x+108-108& =& 126-108\\ 0.60x& =& 18\\ x& =& 30\end{array}$

Step 4. Find the value of y

Substitute $30$for $x$in the third equation

$$\begin{gathered} y=90-x \\ y=90-30 \\ y=60 \end{gathered}$$

So, candy selling is $30$pounds and candy costing is $60$pounds.

Step 5. Check the solution

Substitute $30$for $x$and $60$for $y$in the first equation formed.

$\begin{array}{rcl}x+y& =& 90\\ 30+60& =& 90\\ 90& =& 90\end{array}$

It is a true statement.

Again, substitute the values in the second equation formed.

$\begin{array}{rcl}1.80x+1.20y& =& 126\\ 1.80·30+1.20·60& =& 126\\ 54+72& =& 126\\ 126& =& 126\end{array}$

This is also a true statement.

So the point satisfies both the equations.