First, solve for \( \left|y - 6\right| \):

\( \left|y - 6\right| \) + 3 = 7
\( \left|y - 6\right| \) = 7 - 3
\( \left|y - 6\right| \) = 4

The value inside the absolute value brackets can be either positive or negative so (y - 6) must equal + 4 or -4 for \( \left|y - 6\right| \) to equal 4:

So, y = 2 or y = 10.

The commutative property states that, when adding or multiplying numbers, the order in which they're added or multiplied does not matter. For example, 3 + 4 and 4 + 3 give the same result, as do 3 x 4 and 4 x 3.

To subtract these radicals together their radicands must be the same:

6\( \sqrt{125} \) - 5\( \sqrt{5} \)
6\( \sqrt{25 \times 5} \) - 5\( \sqrt{5} \)
6\( \sqrt{5^2 \times 5} \) - 5\( \sqrt{5} \)
(6)(5)\( \sqrt{5} \) - 5\( \sqrt{5} \)
30\( \sqrt{5} \) - 5\( \sqrt{5} \)

Now that the radicands are identical, you can subtract them:

The first few multiples of 8 are [8, 16, 24, 32, 40, 48, 56, 64, 72, 80] and the first few multiples of 16 are [16, 32, 48, 64, 80, 96]. The first few multiples they share are [16, 32, 48, 64, 80] making 16 the smallest multiple 8 and 16 have in common.

You have 14 out of the total of 350 raffle tickets sold so you have a (\( \frac{14}{350} \)) x 100 = \( \frac{14 \times 100}{350} \) = \( \frac{1400}{350} \) = 4% chance to win the raffle.