Rational Expressions - PSAT Math

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

We will need to simplify the expression $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t}. We can think of this as a large fraction with a numerator of $\frac{1}{t}$-$\frac{1}{x}$ and a denominator of 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. $\frac{1}{t}$ has a denominator of t, and -$\frac{1}{x}$ has a denominator of x. The least common denominator that these two fractions have in common is xt. Thus, we are going to write equivalent fractions with denominators of xt.

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

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

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

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

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

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

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

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

To simplify this further, we can think of $\frac{frac{x-t}{xt}$}{x-t} as the same as $\frac{x-t}{xt}$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, adiv b=acdot $\frac{1}{b}$.

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

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

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

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

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.

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

Add like terms.

Final Answer.

Check for same Denominator

Add like terms

Check for GCF or if the expression can be factored

After factoring, divide out like terms.

Final Answer

We will need to simplify the expression $\frac{(frac{1}{t}$-$\frac{1}{x}$)}{x-t}. We can think of this as a large fraction with a numerator of $\frac{1}{t}$-$\frac{1}{x}$ and a denominator of 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. $\frac{1}{t}$ has a denominator of t, and -$\frac{1}{x}$ has a denominator of x. The least common denominator that these two fractions have in common is xt. Thus, we are going to write equivalent fractions with denominators of xt.

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

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

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

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

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

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

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

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

To simplify this further, we can think of $\frac{frac{x-t}{xt}$}{x-t} as the same as $\frac{x-t}{xt}$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, adiv b=acdot $\frac{1}{b}$.

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

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

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

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

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.

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

Add like terms.

Final Answer.

Check for same Denominator

Add like terms

Check for GCF or if the expression can be factored

After factoring, divide out like terms.

Final Answer

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

Determine an LCD (Least Common Denominator) between and .

LCD =

Multiply the top and bottom of the first rational expression by , so that the denominator will be .

Distribute the to .

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

Final Answer.

Determine an LCD (Least Common Denominator) between and .

LCD =

Multiply the top and bottom of the first rational expression by , so that the denominator will be .

Distribute the to .

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

Final Answer.

If we plug in _a_2 for b in the radical expression, we get √(_a_3) = 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.

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