Question

luke, kira, and ali each served $\frac{2}{3}$ of their own cake. each cake was the same size, but luke served 4 slices, kira served 6 slices, and ali served 8 slices. how is this possible? enter your answers in the boxes to complete the solution.
luke cut his cake into $\frac{4}{6}$ slices, kira cut her cake into $\frac{6}{8}$ slices, and ali cut her cake into $\frac{8}{10}$ slices.

Explanation:

Step1: Solve for Luke's total slices

Let $x$ = total slices of Luke's cake. We know $\frac{2}{3}x = 4$. Rearrange to solve for $x$:
$x = 4 \div \frac{2}{3} = 4 \times \frac{3}{2} = 6$

Step2: Solve for Kira's total slices

Let $y$ = total slices of Kira's cake. We know $\frac{2}{3}y = 6$. Rearrange to solve for $y$:
$y = 6 \div \frac{2}{3} = 6 \times \frac{3}{2} = 9$

Step3: Solve for Ali's total slices

Let $z$ = total slices of Ali's cake. We know $\frac{2}{3}z = 8$. Rearrange to solve for $z$:
$z = 8 \div \frac{2}{3} = 8 \times \frac{3}{2} = 12$
(Note: The given fraction $\frac{8}{10}$ is not a valid total slice count here, as the calculation shows Ali needs 12 total slices to serve $\frac{2}{3}$ as 8 slices.)

Answer:

Luke cut his cake into $\boldsymbol{6}$ slices, Kira cut her cake into $\boldsymbol{9}$ slices, and Ali cut her cake into $\boldsymbol{12}$ slices.

This works because $\frac{2}{3}$ of 6 is 4, $\frac{2}{3}$ of 9 is 6, and $\frac{2}{3}$ of 12 is 8—matching the number of slices each served, even though the total slice counts (and thus individual slice sizes) are different for each cake.