Use PEMDAS (Parentheses, Exponents, Multipy/Divide, Add/Subtract):

3 + (5 + 3) ÷ 4 x 3 - 4²
P: 3 + (8) ÷ 4 x 3 - 4²
E: 3 + 8 ÷ 4 x 3 - 16
MD: 3 + \( \frac{8}{4} \) x 3 - 16
MD: 3 + \( \frac{24}{4} \) - 16
AS: \( \frac{12}{4} \) + \( \frac{24}{4} \) - 16
AS: \( \frac{36}{4} \) - 16
AS: \( \frac{36 - 64}{4} \)
\( \frac{-28}{4} \)
-7

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

4\( \sqrt{112} \) - 6\( \sqrt{7} \)
4\( \sqrt{16 \times 7} \) - 6\( \sqrt{7} \)
4\( \sqrt{4^2 \times 7} \) - 6\( \sqrt{7} \)
(4)(4)\( \sqrt{7} \) - 6\( \sqrt{7} \)
16\( \sqrt{7} \) - 6\( \sqrt{7} \)

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

An integer is any whole number, including zero. An integer can be either positive or negative. Examples include -77, -1, 0, 55, 119.

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

\( \left|y + 6\right| \) + 7 = -7
\( \left|y + 6\right| \) = -7 - 7
\( \left|y + 6\right| \) = -14

The value inside the absolute value brackets can be either positive or negative so (y + 6) must equal - 14 or --14 for \( \left|y + 6\right| \) to equal -14:

So, y = 8 or y = -20.

The first few multiples of 5 are [5, 10, 15, 20, 25, 30, 35, 40, 45, 50] and the first few multiples of 9 are [9, 18, 27, 36, 45, 54, 63, 72, 81, 90]. The first few multiples they share are [45, 90] making 45 the smallest multiple 5 and 9 have in common.