A person wishes to mix coffee worth ​$6 per lb with coffee worth ​$3 per lb to get 180 lb of a mixture worth ​$4 per lb. How many pounds of the ​$6 and the ​$3 coffees will be​ needed?

Answer 1

Set x equal to the amount of $6 coffee

Set y equal to the amount of $3 coffee

Since the amount of total coffee is 180, x+y = 180

Since the total PRICE of the coffee is 180*4 (720), 6x+3y = 720

Use the first equation, x+y = 180, and set it equal to x or y. So x = 180-y or y = 180-x

Plug the equation in respect to x or y into the second equation, 6x+3y = 720 like this: 6(180-y) +3y = 720

OR

6x+3(180-x) = 720

Solve for x or y depending on which equation you plugged in.

Example: 6x + 3(180-x) = 720

x = 60.

Plug x into the original equation, x + y = 180

y = 120.

Therefore: there is 60 coffees that are $6 and 120 coffees that are $3.

Answer 2

Let S = amt of $6 coffee and T = amt of $3 coffee

S + T = 180

6S + 3T = 180*4 => 2S + T = 60*4 = 240

Subtract the first equation from the second

S=60 and then T = 120.