Question

You have one type of nut that sells for $4.40 a pound and another type of nut that sells for $6.10 a pound. You would like to have 8 1/2 pounds of a nut mixture that sells for $5.80 a pound. How much of each not will you need to obtain the desired mixture?

Answer 1

Let x be the number of pounds of the first type ($4.40) of nut, and y be the number of pounds of the other one, then we can set the following system of equations:

$\begin{gathered} x+y=8\frac{1}{2}=8.5, \\ 4.40x+6.10y=5.80\cdot8.50=49.30. \end{gathered}$

Solving the first equation for x we get:

$x=8.5-y\text{.}$

Substituting the above result in the second equation we get:

$\begin{gathered} 4.40(8.5-y)+6.10y=49.30, \\ 37.40-4.40y+6.10y=49.30, \\ 1.7y=11.90, \\ y=7. \end{gathered}$

Substituting y=7 in the third equation on the board we get:

$x=8.5-7=1.5.$

Answer: You will need to add 1.5 pounds of the $4.40 nut and 7 pounds of the other one.