To evaluate, simply plug in the number \(\displaystyle 2\) for \(\displaystyle x\):

\(\displaystyle \frac{2(2^2) + 3}{2}=\)

Remembering to use order of operations (Parentheses, Exponents, Multiplication, Division, Addition, Subtraction) we arrive at our final solution.

\(\displaystyle \frac{2(4)+3}{2}=\frac{8+3}{2}\)

\(\displaystyle =\frac{11}{2}\)

To solve, simply plug in \(\displaystyle 3\) for \(\displaystyle y\):

\(\displaystyle \frac{2y}{y -8}=\)

Remembering to use the correct order of operations (Parentheses, Exponents, Multiplication, Division, Addition, Subtraction) we arrive at our final answer.

First do the multiplication that is in the numerator.

\(\displaystyle \frac{2(3)}{3-8}=\)

Now do the subtraction in the denominator.

\(\displaystyle -\frac{6}{5}\)

To solve, simply plug in \(\displaystyle -1\) for \(\displaystyle x\):

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

Remembering to use the correct order of operations (Parentheses, Exponents, Multiplication, Division, Addition, Subtraction) we arrive at the final solution.

\(\displaystyle \frac{2(-1^2) - 2(-1))}{5} =\)

Also recall that when a negative number is squared it becomes a positive number. This is also true when we multiply two negative numbers together.

\(\displaystyle \frac{2(1)+1}{5}=\)  \(\displaystyle \frac{4}{5}\)

To evaluate, merely plug in \(\displaystyle 4\) for \(\displaystyle x\):

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

\(\displaystyle \frac{4^2 - (3\cdot4)}{4 + 2} =\)

Remembering order of operations (Parentheses, Exponents, Multiplication, Division, Addition, Subtraction) we arrive at our final solution.

\(\displaystyle \frac{16-(12)}{6}=\frac{4}{6}\)

From here we can further reduce the fraction by factoring out a two from both the numerator and denominator.

\(\displaystyle \frac{4}{6} =\frac{2\cdot 2}{2\cdot 3}\)

Canceling out the two's we get:

\(\displaystyle \frac{2}{3}\)

To solve, simply plug in \(\displaystyle 0\) for \(\displaystyle y\):

\(\displaystyle \frac{y^3 - 3}{y - 3}\)

\(\displaystyle \frac{0^3 - 3}{0 - 3}\)

Recall that when a negative number is divided by another negative number it results in a positive number.

\(\displaystyle \frac{-3}{-3}=1\)

To evaluate, simply plug in \(\displaystyle 5\) for \(\displaystyle x\):

\(\displaystyle \frac{x^2 + 2x - 2}{x}\)

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

Remembering the order of operations (Parentheses, Exponents, Multiplication, Division, Addition, Subtraction) we arrive at our final solution.

\(\displaystyle =\frac{25+10-2}{5}=\frac{35-2}{5}=\frac{33}{5}\)

To solve, simply plug in \(\displaystyle -2\) for \(\displaystyle n\):

\(\displaystyle \frac{5n^2}{-2n - 8}\)

\(\displaystyle \frac{5(-2)^2}{(-2\cdot-2)-8}\)

Remembering the order of operations (Parentheses, Exponents, Multiplication, Division, Addition, Subtraction) we arrive at our final solution. Also recall that multiplying two negative numbers together leads to a positive product; this is also true when you square a negative number.

\(\displaystyle \frac{5(4)}{4-8}=\frac{20}{-4}\)

From here we can reduce the fraction by factoring out a four from both the numerator and the denominator.

\(\displaystyle \frac{4\cdot 5}{4 \cdot -1}=-\frac{5}{1}=-5\)

To evaluate, simply plug in \(\displaystyle 3\) for \(\displaystyle z\):

\(\displaystyle \frac{z^2 + 2z - 20}{3}\)

\(\displaystyle \frac{3^2 + (2\cdot 3)-20}{3}\)

Remembering the order of operations we are able to solve this problem. The order of operations is parentheses, exponents, multiplication, division, addition, subtraction.

\(\displaystyle \frac{9 + (6)-20}{3}=\frac{15-20}{3}=-\frac{5}{3}\)

To evaluate, simply plug in \(\displaystyle 2\) for \(\displaystyle x\):

\(\displaystyle \frac{x^3 + x^2}{x}\)

\(\displaystyle \frac{2^3 + 2^2}{2}\)

Recall that we must simplify the exponents before we are able to add the numerator and divide by the denominator.

\(\displaystyle \frac{8 + 4}{2}=\frac{12}{2}\)

From here we can reduce the fraction by factoring out a two from both the numerator and the denominator.

\(\displaystyle \frac{12}{2}=\frac{2\cdot 6}{2\cdot 1}=\frac{6}{1}=6\)

To solve, simply plug in \(\displaystyle -1\) for \(\displaystyle x\):

\(\displaystyle \frac{3x^2 + 9x -18}{2x^2 + x}\)

\(\displaystyle \frac{3(-1)^2 + 9(-1) -18}{2(-1)^2 + -1}\)

According to the order of operations we must simplify the exponents first. Then we will multiply the necessary terms and finally we will add and subtract the terms as specified.

\(\displaystyle =\frac{3(1)-9-18}{2(1)-1}=\frac{3 -27}{2-1}\)

\(\displaystyle =\frac{-24}{1}=-24\)