Rational Expressions

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SAT Math › Rational Expressions

Questions 1 - 10

Function_part1

9/5

–37/15

–11/5

37/15

–9/5

Explanation

Fraction_part2

Fraction_part3

Function_part1

9/5

–37/15

–11/5

37/15

–9/5

Explanation

Fraction_part2

Fraction_part3

Simplify the expression.

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

$\frac{2x}{3x+6}$

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

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

$\frac{1}{2x+4}$

$\frac{x+1}{2x+4}$

Explanation

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.

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

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

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

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

This is the most simplified version of the rational expression.

Simplify the expression.

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

$\frac{2x}{3x+6}$

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

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

$\frac{1}{2x+4}$

$\frac{x+1}{2x+4}$

Explanation

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

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

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

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

This is the most simplified version of the rational expression.

Which of the following is equivalent to $\dpi{100} \frac{(\frac{1}{t}-\frac{1}{x})}{x-t}$ ? Assume that denominators are always nonzero.

$(xt)^{-1}$

$\frac{x}{t}$

$x-t$

$t-x$

$x^{2}-t^{2}$

Explanation

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 $\dpi{100} x-t$ .

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. $\frac{1}{t}$ has a denominator of $\dpi{100} t$ , and $\dpi{100} -\frac{1}{x}$ has a denominator of $\dpi{100} x$ . The least common denominator that these two fractions have in common is $\dpi{100} xt$ . Thus, we are going to write equivalent fractions with denominators of $\dpi{100} xt$ .

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

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

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

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

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, $a\div b=a\cdot \frac{1}{b}$ .

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

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 $(xt)^{-1}$ .

Which of the following is equivalent to $\dpi{100} \frac{(\frac{1}{t}-\frac{1}{x})}{x-t}$ ? Assume that denominators are always nonzero.

$(xt)^{-1}$

$\frac{x}{t}$

$x-t$

$t-x$

$x^{2}-t^{2}$

Explanation

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

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

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

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

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:

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

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

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 $(xt)^{-1}$ .

Simplify (4x)/(x2 – 4) * (x + 2)/(x2 – 2x)

(4x2 + 8x)/(x4 + 8x)

x/(x – 2)2

4/(x – 2)2

x/(x + 2)

4/(x + 2)2

Explanation

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.

Simplify (4x)/(x2 – 4) * (x + 2)/(x2 – 2x)

(4x2 + 8x)/(x4 + 8x)

x/(x – 2)2

4/(x – 2)2

x/(x + 2)

4/(x + 2)2

Explanation

what is 6/8 X 20/3

120/11

18/160

9/40

3/20

Explanation

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

Simplify the following rational expression: (9x - 2)/(x2) MINUS (6x - 8)/(x2)

$\frac{15x+6}{x}$

$\frac{3x+6}{x^{2}}$

$\frac{12x-6}{x}$

Explanation

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.

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