Question

A coffee dealer mixed 12 pounds of one grade coffee with 10 pounds of another grade of coffee to obtain a blend worth $54. He then made a second blend worth $61 by mixing 8 pounds of the first grade with 15 pounds of the second grade. Find the price per pound of each grade.

Answer 1

first grade ---- x lbs
2nd grade ----- y lbs

12x + 10y = 54 ---> 6x+5y = 27
8x + 15y = 61

1st equation times 3
18x + 15y = 81
8x + 15y = 61
subtract them
10x = 20
x = 2
sub back into 6x+5y=27
12+5y=27
y = 3

1st grade is $2 per pound, 2nd grade is $3 per pound

(coffee at $2 a pound!!! WOW, time to update that texbook)

Answer 2

define Variables,
Grade#1=x
Grade#2=y
total cost=z
a=pounds of Grad#1
b=pounds of Grade#2
write equation,
z=ax+by
Define values
12 pounds Grade1 and 10 pounds of Grade2 worth $54; 8 pounds Grade1 and 15 pounds Grade2 worth $61
Plug in values, isolate one variable,
54= 12x+10y , 61= 8x+15y
-12x=-12x , -8x=-8x
54-12x=10y , 61-8x=15y
(54-12x)/10=(10y)/10 , (61-8x)/15=
(54-12x)/10x=y , (15y)/15
(27-6X)/5=y , (61-8x)/15=y
set equations equal to each other, solve for remaining variable,
(27-6x)/5=(61-8x)/15
*15=*15
3(27-6x)=61-8x
81-18x=61-8x
+18x= +18x
81=61+10x
-61=-61
20=10x
(20)/10=(10x)/10
2=y
Pick one equation, Plug in new value, solve for last variable,
54=12x+10y
54=12x+10*(2)
54=12x+20
-20= -20
34=12x
(34)/12=(12x)/12
(17/6)=x
2.83=x

Grade one is $2/lb
Grade two is $2.83/lb