Explaining why cross multiplication works using a pie

Solution 1:

Imagine 9 cakes with 6 pieces each. Both taking 3 of 9 cakes, or 2 pieces of each of the 9 cakes result in 18 pieces of cake.

Solution 2:

Suppose you cut the pie into $6 \cdot 9 = 54$ slices, and call each of those a "unit slice". Then both $2 \cdot 9=3 \cdot 6$ would denote the same amount of "unit slices", which is $\dfrac{18}{54}=\dfrac{1}{3}$ of the full pie.

Without the pie analogy, just note that $\,\dfrac{2}{6} \color{blue}{\cdot \dfrac{9}{9}}=\dfrac{18}{54}=\dfrac{3}{9} \color{blue}{\cdot \dfrac{6}{6}}\,$.

Solution 3:

Imagine 2 cakes with 18 strawberries on top.

1. Take the first cake, cut it into 6, and take 2 slices. (2/6) Each slice has 3 stawberries
2. Take the second cake, cut it into 9 and take 3 slices. (3/9) Each slice has 2 strawberries.

Now, you have 9 slices with 2 strawberries each (9∗2), and 6 slices with 3 strawberries each (6∗3), for a total of 18 strawberries per cake

Solution 4:

Cross multiplication using pie algebra is possible (and tasty), but a little clumsy.

We can read $\frac 26 = \frac 39$ as $$ \text{2 out of 6 slices of pie} = \text{3 out of 9 slices of pie}.\tag{1} $$ To get from $\frac 26 = \frac 39$ to $2\cdot9=3\cdot6$ we need to multiply both sides with six and nine. And we want to do this in terms of pies!

To multiply first by six, we can apply the original identity to six pies at the same time. We get $$ \text{2 out of 6 slices of pie for 6 pies} = \text{3 out of 9 slices of pie for 6 pies} $$ or $\frac 26\cdot6 = \frac 39\cdot6$. Now we can simplify this, because $$ \text{2 out of 6 slices of pie for 6 pies} = \text{2 whole pies}. $$ Instead of cutting each pie in six pieces, you can cut your pile of six pies in six smaller piles (each containing one pie). We have found the cancellation law $\frac26\cdot6=2$.

Now we have $$ \text{2 whole pies} = \text{3 out of 9 slices of pie for 6 pies} $$ and we want to multiply both sides by nine, as we would do algebraically. As before, this involves applying the same identity nine times and tallying the desserts: $$ \text{$2\cdot9$ whole pies} = \text{3 out of 9 slices of pie for $6\cdot9$ pies}. $$ We need to simplify the right-hand side. Multiplying the number of pies by six has the same effect as multiplying the amount of slices by six in the end, so $$ \text{3 out of 9 slices of pie for $6\cdot9$ pies} = (\text{3 out of 9 slices of pie for 9 pies})\cdot6. $$ We can apply the previously used law of pie cancellation to the expression in parentheses to get $$ \text{3 out of 9 slices of pie for 9 pies} = \text{3 whole pies}. $$ We can now conclude that $$ \text{$2\cdot9$ whole pies} = \text{$3\cdot6$ whole pies}.\tag{2} $$ Every step of the reasoning is reversible, so (1) and (2) are equivalent. The numbers $2\cdot9$ and $3\cdot6$ don't represent any natural quantity directly; they arise through algebraic manipulations, whether you use pies or not.

Solution 5:

The first pie has 6 pieces. Divide each piece into 9 smaller pieces. Then 2*9 represents: (number of large pieces taken from first pie) * (number of small pieces in a large piece of the first pie) = (number of small pieces taken from first pie).

Now do the same thing for the second pie, this time splitting each of the 9 pieces into 6 smaller pieces. So 3*6 represents the number of smaller pieces taken from the second pie.

Finally notice that the total number of small pieces is the same in both pies, i.e. each pie is splitted into 6*9=54 small equal pieces. So, assuming the two pies are of the same size, a small piece of the first pie must have the same size with a small piece of the second pie. Thus, if you were to take an equal amount from each pie, you would have to take the same number of small pieces from each pie. I.e. 2*9 = 3*6.