Hello there. To solve this question, we'll have to set up an equation to find the relation between how many gallons Bruce splashed, how many gallons the bathtub holds and how many is left.
Calling the amount of gallons Bruce splashed as the variable g, we know that:
If the bathtub had 18 1/2 gallons of water and after Bruce splashed g gallons of water, there were 8 1/4 left, we'll have that:
[tex]18\frac{1}{2}-g=8\frac{1}{4}[/tex]Add g on both sides of the equation
[tex]g+8\frac{1}{4}=18\frac{1}{2}[/tex]This is the equation we've been looking for.
To find how many gallons Bruce splashed out of the bathtub, subtract 8 1/4 on both sides of the equation
[tex]g=18\frac{1}{2}-8\frac{1}{4}[/tex]Knowing that the notation for mixed fractions:
[tex]a\frac{b}{c}=a+\frac{b}{c}[/tex]We have:
[tex]\begin{gathered} g=18+\frac{1}{2}-\mleft(8+\frac{1}{4}\mright) \\ g=18+\frac{1}{2}-8-\frac{1}{4} \end{gathered}[/tex]Add the fractions
[tex]g=\frac{41}{4}[/tex]Writing this answer in the form of a mixed fraction, we have:
[tex]g=10\frac{1}{4}[/tex]