Pool 1 is a rectangular box 1.3 m deep with a base x metres long and y metres wide. Pool 2 is 1.7 m deep and its base dimensions are each 3 times those of pool 1. Write an expression in x and y for the water needed to fill both pools, then evaluate it at x = 5 and y = 3.

Step-by-step solution

1. 1

   Write the first volume from the box formula. For a rectangular box, volume is length $\times$ width $\times$ depth:

   $V_1=x\cdot y\cdot 1.3=1.3xy$

2. 2

   Scale the second base carefully — this is where the problem bites. Each base dimension is tripled, so the length becomes $3x$ and the width becomes $3y$. The base area is therefore multiplied by $3\times 3=9$, not by $3$:

   $V_2=(3x)(3y)(1.7)=9xy\cdot 1.7=15.3xy$

3. 3

   Add the two volumes.

   $V=V_1+V_2=1.3xy+15.3xy=16.6xy$

   Both terms carry the same variable part $xy$, so they combine like terms directly. This is the answer to part (a).

4. 4

   Substitute the given measurements. With $x=5$ and $y=3$:

   $V=16.6\times 5\times 3=16.6\times 15=249$

5. 5

   Cross-check by computing each pool separately. Pool 1: $5\times 3\times 1.3=19.5$ m³. Pool 2: $15\times 9\times 1.7=135\times 1.7=229.5$ m³. Their sum is

   $19.5+229.5=249\ \text{m}^3\quad\checkmark$

   The two routes agree exactly, confirming $16.6xy$ and the value $249$ m³.