Let B be the number of cups of boiling water that we should mix with I cups of ice to get 3 cups of 110°F water.
Therefore, we can set the following equation:
[tex]\begin{gathered} B+I=3, \\ 220B+32I=110\cdot3. \end{gathered}[/tex]
Subtracting B from the first equation we get:
[tex]\begin{gathered} B+I-B=3-B, \\ I=3-B\text{.} \end{gathered}[/tex]
Substituting the above equation in the second one we get:
[tex]220B+32(3-B)=330.[/tex]
Simplifying the above result we get:
[tex]\begin{gathered} 220B+96-32B=330, \\ 188B+96=330. \end{gathered}[/tex]
Subtracting 96 to the above equation we get:
[tex]\begin{gathered} 188B+96-96=330-96, \\ 188B=234. \end{gathered}[/tex]
Dividing the above equation by 188 we get:
[tex]\begin{gathered} \frac{188B}{188}=\frac{234}{188}, \\ B\approx1.24. \end{gathered}[/tex]
Substituting B=1.24 at I=3-B we get:
[tex]I=3-1.24=1.76.[/tex]
Answer:
[tex]\begin{gathered} 1.26\text{ cups of boiling water.} \\ 1.74\text{ cups of ice.} \end{gathered}[/tex]
