For this division problem, you must deal with the like terms.  

You will divide the constants and then divide the \(\displaystyle x\) variables.  

\(\displaystyle 4\div2=2\) 

and then 

\(\displaystyle x^{3}\div x=x^{2}\) 

because when dividing exponents with common bases, you just subtract the exponents.  

The \(\displaystyle y\) variable remains unchanged and your answer is,

\(\displaystyle 2x^{2}y^{2}\).

To solve  

\(\displaystyle 7^{4} \div (18-11) =\)

First, use the order of operations.

\(\displaystyle 7^{4} \div 7 =\)

When dividing with variables and the coefficients are the same, subtract the exponents.

\(\displaystyle 7^{4} \div 7^{1} = 7^{4-1}\)

\(\displaystyle 7^{4-1} = 7^{3}\)\(\displaystyle 7^{3} = 7\times 7\times 7 = 343\)

To solve, subtract the exponents

 \(\displaystyle \frac{x^{8}}{x^{2}} = x^{8-2}\)

\(\displaystyle x^{6}\)

To solve \(\displaystyle \frac{n^{3}}{n^{6}}\) subtract the exponents

\(\displaystyle n^{3-6} = n^{-3}\)

\(\displaystyle n^{-3} = n\div -3 =\)

\(\displaystyle \frac{1}{n^{3}}\)

The formula for the Area of a rectangle is:

A - length x width

Is this problem, you are given the amount of total area or A, which is\(\displaystyle 36ab^{3}\) square units, and you are given the measurement of the length, which is \(\displaystyle 6ab\).  In order to solve for the measurement of the width, divide.  When dividing, exponents are subtracted.

\(\displaystyle 36\div6 =6\)

\(\displaystyle \frac{a^{1}}{a^{1}} = a^{1-1} = a^{0} = 1\)

\(\displaystyle \frac{b^{3}}{b^{1}} = b^{3-1} = b^{2}\)

\(\displaystyle 6 (1) \times b^{2} = 6b^{2}\) units 

To solve \(\displaystyle \frac{9x^{4}y^{6}}{3x^{2}y^{2}}\) separate each part of the terms and divide.  When dividing variables with exponents, subtract the exponents.

\(\displaystyle 9\div 3 = 3\)

\(\displaystyle \frac{x^{4}}{x^{2}}=x^{4-2} =x^{2}\)

\(\displaystyle \frac{y^{6}}{y^{2}}=y^{6-2} =y^{4}\)

\(\displaystyle 3x^{2}y^{4}\)

Separate each part of the division problem. When dividing, exponents are subtracted.

\(\displaystyle 9 \div3 =3\)

\(\displaystyle \frac{x^{5}}{x^{3}} = x^{5-3} = x^{2}\)

\(\displaystyle \frac{y^{4}}{y^{9}} = y^{4-9} = y^{-5}\)

\(\displaystyle 3x^{2}y^{-5} =\)

\(\displaystyle \frac{3x^{2}}{y^{5}}\)

Simplify the following statement.

\(\displaystyle \frac{14x^6}{7x^3}\)

TO divide variables, we need to subtract the exponents. The coefficients (numbers out in front) can be divided as you would normally divide.

\(\displaystyle \frac{14x^6}{7x^3}=\frac{14}{7}*x^{6-3}=2x^3\)

So we get our answer of two x cubed.

\(\displaystyle 14x - 5 (x + 8)\)

\(\displaystyle = 14x - 5 \cdot x - 5 \cdot 8\)

\(\displaystyle = 14x - 5x - 40\)

\(\displaystyle = (14- 5) x - 40\)

\(\displaystyle = 9x - 40\)

\(\displaystyle 8 (x - 7) - 3(x + 2)\)

\(\displaystyle = 8 \cdot x -8 \cdot 7 - 3 \cdot x + (-3) \cdot 2\)

\(\displaystyle = 8x -56 - 3 x -6\)

\(\displaystyle = 8x - 3 x -56 -6\)

\(\displaystyle =( 8 - 3 ) x - (56 + 6)\)

\(\displaystyle =5 x - 62\)