The distributive property for division helps in solving expressions like \({b + c \over a}\). It specifies that the result of dividing a fraction with multiple terms in the numerator and one term in the denominator can be obtained by dividing each term individually and then totaling the results: \({b + c \over a} = {b \over a} + {c \over a}\). For example, \({a^3 + 6a^2 \over a^2} = {a^3 \over a^2} + {6a^2 \over a^2} = a + 6\).

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

5 + (4 + 3) ÷ 3 x 5 - 4²
P: 5 + (7) ÷ 3 x 5 - 4²
E: 5 + 7 ÷ 3 x 5 - 16
MD: 5 + \( \frac{7}{3} \) x 5 - 16
MD: 5 + \( \frac{35}{3} \) - 16
AS: \( \frac{15}{3} \) + \( \frac{35}{3} \) - 16
AS: \( \frac{50}{3} \) - 16
AS: \( \frac{50 - 48}{3} \)
\( \frac{2}{3} \)
\(\frac{2}{3}\)

To add or subtract terms with exponents, both the base and the exponent must be the same. In this case they are so add the coefficients and retain the base and exponent:

-6x² + 1x²
(-6 + 1)x²
-5x²

To multiply fractions, multiply the numerators together and then multiply the denominators together:

\( \frac{2}{9} \) x \( \frac{4}{7} \) = \( \frac{2 x 4}{9 x 7} \) = \( \frac{8}{63} \) = \(\frac{8}{63}\)

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

9\( \sqrt{32} \) + 9\( \sqrt{2} \)
9\( \sqrt{16 \times 2} \) + 9\( \sqrt{2} \)
9\( \sqrt{4^2 \times 2} \) + 9\( \sqrt{2} \)
(9)(4)\( \sqrt{2} \) + 9\( \sqrt{2} \)
36\( \sqrt{2} \) + 9\( \sqrt{2} \)

Now that the radicands are identical, you can add them together: