SOLUTION: I set up a table for this question, but the numbers I got were too big to be right. Here is the problem: A candy mix is being prepared by mixing candy that is worth $12 per

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Question 24528: I set up a table for this question, but the numbers I got were too big to be right.
Here is the problem:
A candy mix is being prepared by mixing candy that is worth $12 per pound with candy that is worth $15 per pound. The candy store wants to have 120 pounds of this mixture which will sell for $13.75 per pound. How many pounds of each type of candy should be used?
Find the number of pounds @$12 and @$15.
I made a table with x and y representing the # of lbs. With 120lbs. as the total. 12 and 15 as the price per pound and 13.75 for the total price per pound.
In the third boxes of my table I wrote 12x, 15x, and 120(13.75)=1650.
Then I wrote the equation: 12x+15y=1650
(here is where I get lost)
I then did: 12(120-y)+y=1650
= 1440-12y+y=1650 = 1440-11y=1650
I subtracted 1440 from both sides to isolate the variable and came up with:
-11y=210
then I divided -11y and 210 by -11= y= -19.09
and then I solved for x by: 120-19.09= 100.91 x=100.91!!!!
I KNOW THIS IS TOTALLY WRONG, BUT THERE IS NOT ANYTHING IN MY TEXT BOOK TO SHOW ME HOW TO SOLVE THIS PROBLEM!!!! PLEASE HELP!! THIS HOMEWORK IS DUE BY 01/17/06
HELP WILL BE GREATLY APPRECIATED! :)
Answer by Earlsdon(6294) (Show Source):
You can put this solution on YOUR website!
Try this:
Let x = the required number of pounds at $12.00 a pound (12x), then (120-x) = the number of pounds at $15.00 a pound 15(120-x). The sum of these are to equal 120 pounds at $13.75 a pound (120(13.75). Let's set up the equation to find x.
Simplify and solve for x.
Combine like-terms.
Add 3x to both sides of the equation.
Subtract 1650 from both sides.
Now divide both sides by 3.

The candy store will have to mix 50 pounds of $12.00 candy with 120-50= 70 pounds of $15.00 candy to obtain 120 pounds of $13.75 candy.