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.

To simplify the following, a common denominator must be achieved. In this case, the first term must be multiplied by (x+2) in both the numerator and denominator and likewise with the second term with (x-3).

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

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

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. 

To solve this, we need to set a proportion.

\(\displaystyle \frac{100\:\text{ft}}{20\sec}=\frac{1500\:\text{ft}}{x}\)

Cross Multiply

\(\displaystyle 100\cdot x=1500\cdot 20\)

\(\displaystyle x=\frac{30000}{100}=300\)

So it will take Jill \(\displaystyle 300 \sec\) to walk \(\displaystyle 1500\: \text{ft}\)

\(\displaystyle \frac{b+c}{c}=\frac{14}{15}\)

\(\displaystyle \frac{b}{c}+\frac{c}{c}=\frac{14}{15}\)

\(\displaystyle \frac{b}{c}+1=\frac{14}{15}\)

\(\displaystyle \frac{b}{c}+1\ {\color{Red} -1}=\frac{14}{15}\ {\color{Red} -1}\)

\(\displaystyle \frac{b}{c}=\frac{14}{15}-\frac{15}{15}\)

\(\displaystyle \frac{b}{c}=-\frac{1}{15}\)

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

The procedure for multplying together two rational expressions is the same as multiplying together any two fractions: find the product of the numerators and the product of the denominators separately, and then simplify the resulting quotient as far as possible, as shown:

\(\displaystyle \frac{15x^2+4xy^3-3y}{x^2y^7}\times\frac{5y^4}{4x^3}\)

\(\displaystyle =\frac{(15x^2+4xy^3-3y)(5y^4)}{(x^2y^7)(4x^3)}\)

\(\displaystyle =\frac{75x^2y^4+20xy^7-15y^5}{4x^5y^7}\)

\(\displaystyle =\frac{75x^2+20xy^3-15y}{4x^5y^3}\)