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.
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)
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
To find the price per pound of each grade of coffee, we can set up a system of linear equations. Let's assign variables to the unknowns:
Let x be the price per pound of the first grade coffee.
Let y be the price per pound of the second grade coffee.
Now, let's set up the equations based on the given information:
First equation:
12x + 10y = 54
This equation represents the value of the blend when 12 pounds of the first grade coffee is mixed with 10 pounds of the second grade coffee, resulting in a blend worth $54.
Second equation:
8x + 15y = 61
This equation represents the value of the second blend when 8 pounds of the first grade coffee is mixed with 15 pounds of the second grade coffee, resulting in a blend worth $61.
Now that we have the system of equations:
12x + 10y = 54
8x + 15y = 61
We can solve this system to find the values of x and y.
There are various methods to solve a system of linear equations, such as substitution, elimination, or using matrices. Let's use the substitution method.
1. Solve the first equation for x:
12x = 54 - 10y
x = (54 - 10y) / 12
2. Substitute the value of x in the second equation:
8((54 - 10y) / 12) + 15y = 61
Now, we can solve this equation algebraically to find the value of y.
8(54 - 10y) + 180y = 732 (multiplied both sides by 12 to get rid of the fraction)
432 - 80y + 180y = 732 (distributed 8 across the brackets)
100y = 300 (combined like terms)
y = 300 / 100 = 3
Now that we have the value of y, we can substitute it back into the first equation to find the value of x:
x = (54 - 10(3)) / 12
x = (54 - 30) / 12
x = 24 / 12 = 2
Therefore, the price per pound of the first grade coffee is $2, and the price per pound of the second grade coffee is $3.