Question

You have one type of chocolate that sells for $2.50/1b and another type of chocolate that sells for $8.50/lb. You would like to have 36 lbs of a chocolate mixture that sells for $6.40/1b. How much of each chocolate will you need to obtain the desired mixture? You will need lbs of the cheaper chocolate and lbs of the expensive chocolate.

Answer 1

We have the next information

First type of chocolate $2.50 per lb

Second type of choclate 8.50 per lb

x= lbs of the cheaper chocolate

y=lbs of the expensive chocolate

36 total pounds

x+y=36

2.50(x)+8.50(y)=36(6.40)

2.5x+8.5y=230.4

then we solve the equation system

x+y=36 .....(1)

2.5x+8.5y=230.4 ....(2)

we clear x from the first equation

x=36-y

we substitute the equation above in the second equation

2.5(36-y)+8.5y=230.4

90-2.5y+8.5y=230.4

we clear y

6y=230.4-90

6y=140.4

y=140.4/6

y=23.4

then we substitute the value of y in the next equation

x=36-y=36-23.4

x=12.6

you will need 12.6 lbs of the cheaper chocolate and 23.4 lbs of the expensive chocolate