$\displaystyle \frac{2x-3}{x+2}+\frac{5x-6}{x+2}$

Same denominator means you add straight across the numerators, keeping the denominator the same.

$\displaystyle \frac{2x-3+5x-6}{x+2}$

Add like terms. 

$\displaystyle \frac{2x+5x+3-6}{x+2}$

Final Answer. 

$\displaystyle \frac{7x-3}{x+2}$

$\displaystyle \frac{x^{2}-4}{2x+8}+\frac{-12}{2x+8}$

Check for same Denominator

$\displaystyle \frac{x^{2}-4+(-12)}{2x+8}$

Add like terms 

$\displaystyle \frac{x^{2}-16}{2x+8}$

Check for GCF or if the expression can be factored

$\displaystyle \frac{(x-4)(x+4)}{2(x+4)}$

After factoring, divide out like terms. $\displaystyle x+4$

$\displaystyle \frac{(x-4)}{2}$

Final Answer

Since both expressions have a common denominator, x², we can just recopy the denominator and focus on the numerators. We get (9x - 2) - (6x - 8). We must distribute the negative sign over the 6x - 8 expression which gives us 9x - 2 - 6x + 8 ( -2 minus a -8 gives a +6 since a negative and negative make a positive). The numerator is therefore 3x + 6.

To add rational expressions, first find the least common denominator. Because the denominator of the first fraction factors to 2(x+2), it is clear that this is the common denominator. Therefore, multiply the numerator and denominator of the second fraction by 2.

$\displaystyle \frac{x}{2x+4}+\frac{x}{x+2}$

$\displaystyle \frac{x}{2x+4}+\frac{(2)x}{(2)(x+2)}$

$\displaystyle \frac{x}{2x+4}+\frac{2x}{2x+4}$

$\displaystyle \frac{3x}{2x+4}$

This is the most simplified version of the rational expression.

If we plug in a² for b in the radical expression, we get √(a³) = 8. This can be rewritten as a^3/2 = 8. Thus, log_a 8 = 3/2. Plugging in the answer choices gives 4 as the correct answer. 

$\displaystyle \frac{x+5}{x}-\frac{2x-3}{x^{2}}$

Determine an LCD (Least Common Denominator) between $\displaystyle x$ and $\displaystyle x^{2}$.

LCD = $\displaystyle x^{2}$

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

Multiply the top and bottom of the first rational expression by $\displaystyle x$, so that the denominator will be $\displaystyle x^{2}$.

Distribute the $\displaystyle x$ to $\displaystyle x+5$.

$\displaystyle \frac{x^{2}+5x}{x^{2}}-\frac{2x-3}{x^{2}}$

Now you can subtract because both rational expressions have the same denominators.

$\displaystyle \frac{x^{2}+5x-2x+3}{x^{2}}$

Final Answer. 

$\displaystyle \frac{x^{2}+3x+3}{x^{2}}$

We will need to simplify the expression $\displaystyle \frac{(\frac{1}{t}-\frac{1}{x})}{x-t}$. We can think of this as a large fraction with a numerator of $\displaystyle \frac{1}{t}-\frac{1}{x}$ and a denominator of $\displaystyle \dpi{100} x-t$.

In order to simplify the numerator, we will need to combine the two fractions. When adding or subtracting fractions, we must have a common denominator. $\displaystyle \frac{1}{t}$ has a denominator of $\displaystyle \dpi{100} t$, and $\displaystyle \dpi{100} -\frac{1}{x}$ has a denominator of $\displaystyle \dpi{100} x$. The least common denominator that these two fractions have in common is $\displaystyle \dpi{100} xt$. Thus, we are going to write equivalent fractions with denominators of $\displaystyle \dpi{100} xt$.

In order to convert the fraction $\displaystyle \dpi{100} \frac{1}{t}$ to a denominator with $\displaystyle \dpi{100} xt$, we will need to multiply the top and bottom by $\displaystyle \dpi{100} x$.

$\displaystyle \frac{1}{t}=\frac{1\cdot x}{t\cdot x}=\frac{x}{xt}$

Similarly, we will multiply the top and bottom of $\displaystyle \dpi{100} -\frac{1}{x}$ by $\displaystyle \dpi{100} t$.

$\displaystyle \frac{1}{x}=\frac{1\cdot t}{x\cdot t}=\frac{t}{xt}$

We can now rewrite $\displaystyle \frac{1}{t}-\frac{1}{x}$ as follows:

$\displaystyle \frac{1}{t}-\frac{1}{x}$ = $\displaystyle \frac{x}{xt}-\frac{t}{xt}=\frac{x-t}{xt}$

Let's go back to the original fraction $\displaystyle \frac{(\frac{1}{t}-\frac{1}{x})}{x-t}$. We will now rewrite the numerator:

$\displaystyle \frac{(\frac{1}{t}-\frac{1}{x})}{x-t}$ = $\displaystyle \frac{\frac{x-t}{xt}}{x-t}$

To simplify this further, we can think of $\displaystyle \frac{\frac{x-t}{xt}}{x-t}$ as the same as $\displaystyle \frac{x-t}{xt}\div (x-t)$ . When we divide a fraction by another quantity, this is the same as multiplying the fraction by the reciprocal of that quantity. In other words, $\displaystyle a\div b=a\cdot \frac{1}{b}$.

$\displaystyle \frac{x-t}{xt}\div (x-t)$ = $\displaystyle \frac{x-t}{xt}\cdot \frac{1}{x-t}$$\displaystyle =\frac{x-t}{xt(x-t)}$$\displaystyle = \frac{1}{xt}$

Lastly, we will use the property of exponents which states that, in general, $\displaystyle \frac{1}{a}=a^{-1}$.

$\displaystyle \frac{1}{xt}=(xt)^{-1}$

The answer is $\displaystyle (xt)^{-1}$.

Factor first.  The numerators will not factor, but the first denominator factors to (x – 2)(x + 2) and the second denomintaor factors to x(x – 2).  Multiplying fractions does not require common denominators, so now look for common factors to divide out.  There is a factor of x and a factor of (x + 2) that both divide out, leaving 4 in the numerator and two factors of (x – 2) in the denominator.

6/8 X 20/3 first step is to reduce 6/8 -> 3/4 (Divide top and bottom by 2)

3/4 X 20/3 (cross-cancel the threes and the 20 reduces to 5 and the 4 reduces to 1)

1/1 X 5/1 = 5