A man bought 2 pounds of coffee and 1 pound of butter for a total of $11.25. A month later, the prices had not changed (this makes it a fictitious problem), and he bought 3 pounds of coffee and 2 pounds of butter for $18.50. Find the price per pound of both the coffee and the butter.

See explanation

Step-by-step explanation:

Let the price per pound of coffee be c

Let the price ce per pound of butter be b.

We can form an equation based on the information given and this will be:

2c + b = 11.25 i

3c + 2b = 18.50 ii

From equation i, b = 11.25 - 2c iii

Put the value of b into equation ii.

3c + 2b = 18.50

3c + 2(11.25 - 2c) = 18.50

3c + 22.50 - 4c = 18.50

-c = 18.50 - 22.50

c = 4.00

A pound of coffee cost $4.00

Since b = 11.25 - 2c .

b = 11.25 - 2(4.00)

b = 11.25 - 8.00

b = 3.25

A pound of butter costs $3.25

2 x^2 -3x+p

sum of roots = 3/2

let other root be x

3+ x = 3/2

x= 3/2 - 3 = -3/2

product of roots = p/2

3× -3/2 = p/2

-9 = p

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