In order to determine how many pounds of $7.28 coffe should Mr. May mix with 9 pounds of the $8.40 coffee, you first write the given situation in an algebraic way.
If x is the needed number of pounds $7.28 coffee, 7.28x is the cost of x pounds of such a coffe. This cost, added to the cost of the 9 pounds of $8.40 coffee is equal to y (total cost of both type of coffe).
Then you have:
7.28x + 9(8.40) = y
this is the total price for x pounds of $7.28 coffe and 9 pounds of $8.40.
y/(x + 9) = 7.95
this is the price per pound of the mixed coffe.
Then, you have the following system of equations:
y/(x + 9) = 7.95
7.28x + 9(8.40) = y
To solve for y and x, you proceed as follow:
solve the first equation for y:
y/(x + 9) = 7.95
y = 7.95(x + 9)
y = 7.95x + 71.55
next, you equal the previous equation with the equation 7.28x + 9(8.40) = y, and you solve for x:
7.95x + 71.55 = 7.28x + 9(8.40)
7.95x + 71.55 = 7.28x + 75.6 subtract 7.28x both sides
7.95x - 7.28x + 71.55 = 75.6 subtract 71.55 both sides
7.95x - 7.28x = 75.6 - 71.55 simplify like terms
0.67x = 4.05 divide by 0.67 both sides
x = 4.05/0.67
x = 6.04 ≈ 6
next, you replace the previous value of x in one of the equations:
y = 7.95x + 71.55
y = 7.95(6) + 71.55 = 119.25
Hence, the number of pounds of $7.28 coffee should Mr. May mix with the 9 pounds of $8.40 is 6 pounds.