Kyle works at a donut​ factory, where a​ 10-oz cup of coffee costs 95¢​, a​ 14-oz cup costs​ $1.15, and a​ 20-oz cup costs​ $1.50. During one busy​ period, Kyle served 21 cups of​ coffee, using 294 ounces of​ coffee, while collecting a total of ​$24.35. How many cups of each size did Kyle​ fill? Kyle filled ___ 10-oz cup(s), ___ 14-oz cup(s), and ___ ​20-oz cup(s).

kyle works at a donut shop, where a 10oz. cup of coffee costs 95c, a 14 oz. cup of coffee costs $1.15, and a 20oz. cup costs $1.50. during one busy period Kyle served 24 cups of coffee, using 384 ounces of coffee, while collecting a total of $30.60. How many cups of each size did kyle fill.

Let x = no. of 10 oz cups sold

Let y = no. of 14 oz cups sold

Let z = no. of 20 oz cups sold

Equation 1: total number of cups sold:

x + y + z = 24

Equation 2: amt of coffee consumed:

10x + 14y + 20z = 384

Equation 3: total revenue from cups sold

.95x + 1.15y + 1.50z = 30.60

Mult the 1st equation by 20 and subtract the 2nd equation from it:

20x + 20y + 20z = 480

10x + 14y + 20z = 384

subtracting eliminates z

10x + 6y = 96; (eq 4)

Mult the 1st equation by 1.5 and subtract the 3rd equation from it:

1.5x + 1.5y + 1.5z = 36.00

.95x + 1.15y+ 1.5z = 30.60

subtracting eliminates z again

.55x + .35y = 5.40; (eq 5)

Multiply eq 4 by .055 and subtract from eq 5:

.55x + .35y = 5.40

.55x + .33y = 5.28

eliminates x

0x + .02y = .12

y = .12/.02

y = 6 ea 14 oz cups sold

Substitute 6 for y for in eq 4

10x + 6(6) = 96

10x = 96 - 36

x = 60/10

x = 6 ea 10 oz cups

That would leave 12 ea 20 oz cups (24 - 6 - 6 = 12)

Check our solutions in eq 2:

10(6) + 14(6) + 20(12) =

60 + 84 + 240 = 384 oz

only 10/60 isn't equivalent

hope this : )

80x^8−68x^6+16^5−20x^4−4x^3+8x^2

hope i'm right lol