Answer:
Kyle filled 4 10-oz cups, 6 14-oz cups, and 4 20-oz cups.
Step-by-step explanation:
Let 10-oz, 14-oz, and 20-oz coffees be represented by the variables a, b, and c, respectively.
Since a total of 14 cups of coffee was served:
[tex]a+b+c=14[/tex]
A total of 204 ounces of coffee was served. Therefore:
[tex]10a+14b+20c=204[/tex]
A total of $16.70 was collected. Hence:
[tex]0.95a+1.15b+1.5c=16.7[/tex]
This yields a triple system of equations. In order to solve a triple system, we should isolate the system to only two variables first.
From the first equation, let's subtract a and b from both sides:
[tex]c=14-a-b[/tex]
Substitute this into both the second and third equations:
[tex]10a+14b+20(14-a-b)=204[/tex]
And:
[tex]0.95a+1.15b+1.5(14-a-b)=16.7[/tex]
In this way, we've successfully created a system of two equations, which can be more easily solved. Distribute:
For the Second Equation:
[tex]\displaystyle \begin{aligned} 10a+14b+280-20a-20b&=204\\ -10a-6b&=-76\\5a+3b&=38\end{aligned}[/tex]
And for the Third:
[tex]\displaystyle \begin{aligned} 0.95a+1.15b+21-1.5a-1.5b&=16.7\\ -0.55a-0.35b&=-4.3\end{aligned}[/tex]
We can solve this using substitution. From the second equation, isolate a:
[tex]\displaystyle a=\frac{1}{5}(38-3b)=7.6-0.6b[/tex]
Substitute into the third:
[tex]-0.55(7.6-0.6b)-0.35b=-4.3[/tex]
Distribute and simplify:
[tex]-4.18+0.33b-0.35b=-4.3[/tex]
Therefore:
[tex]-0.02b=-0.12\Rightarrow b=6[/tex]
Using the equation for a:
[tex]a=7.6-0.6(6)=4[/tex]
And using the equation for c:
[tex]c=14-(4)-(6)=14-10=4[/tex]
Therefore, Kyle filled 4 10-oz cups, 6 14-oz cups, and 4 20-oz cups.
