SOLUTION: A dealer mixed some coffee worth $.80 per pound with coffee worth $.50 per pound to make a mixture to be sold for $.70 per pound. If the number of pounds of $.80 coffee was 10 more

Question 124567: A dealer mixed some coffee worth $.80 per pound with coffee worth $.50 per pound to make a mixture to be sold for $.70 per pound. If the number of pounds of $.80 coffee was 10 more than the number of pounds of the $.50 coffee, how many pounds of each kind did he use?
Found 2 solutions by Earlsdon, checkley71:
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Start by letting x = the number of pounds of the $0.50 per pound coffee.
The the number of pounds of the $0.08 per pound coffee can be expressed by (x+10)
The sum of these two quantities (x+x+10)is to equal the number of pounds of coffee at $0.70 per pound.
We can write the following equation:
Simplify this and solve for x.

subtract 1.3x from both sides.
Subtract 7 from both sides.
Divide both sides by 0.1

The dealer will need to mix 10 pounds of the $0.50 per pound coffee (x+10) 20 pounds of the $0.80 per pound coffee to obtain 30 pounds of $0.70 per pound coffee.
Answer by checkley71(8403) (Show Source): You can put this solution on YOUR website!
.8(x+10)+.5x=.7(2x+10)
.8x+8+.5x=1.4x+7
.8x+.5x-1.4x=7-8
-.1x=-1
x=-1/-.1
x=10 pounds of $.50 coffee is used.
10+10=20 pounds of $.80 coffee is used.
proof
.8*20+.5*10=.7*30
16+5=21
21=21

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