Rational Expressions - Algebra 2

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Question

Subtract the following expressions: $\frac{x}{2x+1}-\frac{1}{2x}$

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Answer

In order to subtract the fractions, multiply both denominators together in order to obtain the least common denominator.

$\frac{x}{2x+1}-\frac{1}{2x} = \frac{x(2x)}{(2x+1)(2x)}-\frac{1(2x+1)}{(2x+1)(2x)}$

Simplify the numerators.

$= \frac{2x^2}{(2x+1)(2x)}-\frac{2x+1}{(2x+1)(2x)}$

Combine the numerators.

$=\frac{2x^2-(2x+1)}{(2x+1)(2x)}=\frac{2x^2-2x-1}{4x^2+2x}$

The answer is: $\frac{2x^2-2x-1}{4x^2+2x}$

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Question

Subtract: $\frac{3}{2-x}-\frac{5}{2x}$

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Answer

Multiply the denominators to get the least common denominator. We can then convert both fractions so that the denominators are alike.

$\frac{3}{2-x}-\frac{5}{2x}=\frac{3(2x)}{(2-x)(2x)}-\frac{5(2-x)}{2x(2-x)}$

Simplify both the top and the bottom.

$\frac{6x}{-2x^2+4x} - \frac{10-5x}{-2x^2+4x}$

Combine the numerators as one fraction. Be careful with the second fraction since the entire numerator is a quantity, which means we will need to brace with parentheses.

$\frac{6x-(10-5x)}{-2x^2+4x} =\frac{6x-10+5x}{-2x^2+4x} = \frac{11x-10}{-2x^2+4x}$

Pull out a common factor of negative one in the denominator. This allows us to rewrite the fraction with the negative sign in front of the fraction.

$\frac{11x-10}{-2x^2+4x}= \frac{11x-10}{-(2x^2-4x)}= -\frac{11x-10}{2x^2-4x}$

The answer is: $-\frac{11x-10}{2x^2-4x}$

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Question

Solve: $\frac{2}{x+3}+\frac{8}{x}$

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Answer

To simplify this expression, we will need to multiply both denominators together to find the least common denominator.

Convert both fractions to the common denominator.

$\frac{2}{x+3}+\frac{8}{x}=\frac{2x}{x^2+3x}+\frac{8(x+3)}{x^2+3x}$

Combine the fractions.

$\frac{2x+8x+24}{x^2+3x} = \frac{10x+24}{x^2+3x}$

The answer is: $\frac{10x+24}{x^2+3x}$

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Question

Add: $\frac{7}{9x}+\frac{2(4-x)}{x}$

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Answer

Evaluate by changing the denominator of the second fraction so that both fractions have similar denominators.

$\frac{7}{9x}+\frac{2(4-x)}{x} = \frac{7}{9x}+\frac{2[9](4-x)}{[9]x}$

Use distribution to simplify the numerator.

$\frac{7}{9x}+\frac{2[9](4-x)}{[9]x} =\frac{7}{9x}+\frac{72-18x}{9x}$

Combine like-terms.

The answer is: $\frac{-18x+79}{9x}$

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Question

Add: $\frac{3}{10x}-\frac{10}{x}$

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Answer

In order to subtract the fractions, we will need to determine the least common denominator.

Multiply the second fraction's numerator and denominator by ten, since both fractions share an x-term.

$\frac{3}{10x}-\frac{10(10)}{x(10)} = \frac{3}{10x}-\frac{100}{10x}$

Combine the numerator to form one fraction.

The answer is: $-\frac{97}{10x}$

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Question

Solve: $\frac{x}{2x-7}+ \frac{7x}{x-7}$

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Answer

In order to add the numerators, we will need to find the least common denominator.

Use the FOIL method to multiply the two denominators together.

Simplify by combining like-terms. The LCD is:

Change the fractions.

$\frac{x(x-7)}{(2x-7)(x-7)}+ \frac{7x(2x-7)}{(x-7)(2x-7)}$

Simplify the fractions.

$\frac{x^2-7x}{2x^2-21x+49}+\frac{14x^2-49x}{2x^2-21x+49}$

Combine like-terms.

The answer is: $\frac{15x^2-56x}{2x^2-21x+49}$

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Question

Add: $\frac{3}{2x}+\frac{4}{3x}+\frac{5}{4x}$

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Answer

Determine the least common denominator in order to add the numerator.

Each denominator shares an term. The least common denominator is since it is divisible by each coefficient of the denominator.

Convert the fractions.

$\frac{3}{2x}+\frac{4}{3x}+\frac{5}{4x} = \frac{3(6)}{2x(6)}+\frac{4(4)}{3x(4)}+\frac{5(3)}{4x(3)}$

Simplify the top and bottom.

$\frac{18}{12x}+\frac{16}{12x}+\frac{15}{12x}=\frac{49}{12x}$

The answer is: $\frac{49}{12x}$

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Question

Solve for .

$\frac{x}{4}=\frac{3}{4}$

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Answer

To solve for the variable , isolate the variable on one side of the equation with all other constants on the other side. To accomplish this perform the opposite operation to manipulate the equation.

First cross multiply.

$\frac{x}{4}=\frac{3}{4}$

Next, divide by four on both sides.

$\frac{4x}{4}=\frac{12}{4}=\frac{4\cdot 3}{4}=3$

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Question

Solve: $\frac{7}{5x}+\frac{9}{10x}$

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Answer

In order to solve this expression, we will need to determine the least common denominator. Notice that the denominators both share an x term. We do not need to change that.

Multiply the denominator of the first fraction by two to get the least common denominator, which is .

$\frac{7(2)}{5x(2)}+\frac{9}{10x} = \frac{14}{10x}+\frac{9}{10x}$

Add the numerators.

The answer is: $\frac{23}{10x}$

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Question

Solve: $\frac{9x}{x+1}-\frac{9}{x}$

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Answer

Determine the least common denominator by multiplying the denominators together.

Convert the fractions given.

$\frac{9x}{x+1}-\frac{9}{x}=\frac{(x)(9x)}{(x)(x+1)}-\frac{9(x+1)}{x(x+1)}$

Simplify the numerators and denominators.

$\frac{9x^2}{x^2+x}-\frac{9x+9}{x^2+x}$

Combine the terms as one fraction. Make sure to brace the term since this is a quantity.

$\frac{9x^2}{x^2+x}-\frac{9x+9}{x^2+x} = \frac{9x^2-(9x+9)}{x^2+x}$

The answer is: $\frac{9x^2-9x-9}{x^2+x}$

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Question

Add: $\frac{4x}{x-1}-\frac{10}{x}$

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Answer

Identify the least common denominator by multiplying the denominators together.

Convert the fractions.

$\frac{4x}{x-1}-\frac{10}{x} = \frac{4x(x)}{x(x-1)}-\frac{10(x-1)}{x(x-1)}$

Simplify the numerator and denominator.

$\frac{4x^2}{x^2-x}-\frac{10x-10}{x^2-x}$

Combine both fractions as one. Make sure to enclose the second number in parentheses since the negative sign is distributive.

$\frac{4x^2}{x^2-x}-\frac{10x-10}{x^2-x}=\frac{4x^2-(10x-10)}{x^2-x} =\frac{4x^2-10x+10}{x^2-x}$

The answer is: $\frac{4x^2-10x+10}{x^2-x}$

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Question

$\frac{15x+A}{x^{2}+8x+15}=\frac{7}{x+5}+\frac{8}{x+3}$

Determine the value of .

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Answer

(x+5)(x+3) is the common denominator for this problem making the numerators 7(x+3) and 8(x+5).

7(x+3)+8(x+5)= 7x+21+8x+40= 15x+61

A=61

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Question

Which of the following fractions is NOT equivalent to $- \frac{x-5}{2x + 3}$ ?

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Answer

We know that $-\frac{a}{b}$ is equivalent to $\frac{-a}{b}$ or $\frac{a}{-b}$ .

By this property, there is no way to get $\frac{x+5}{2x+3}$ from $-\frac{x-5}{2x+3}$ .

Therefore the correct answer is $\frac{x+5}{2x+3}$ .

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Question

Simplify:

$\frac{x^2-1}{x^2+4x+2}\div\frac{x^2+2x+1}{x^2+3x+2}$

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Answer

This problem is a lot simpler if we factor all the expressions involved before proceeding:

$\frac{x^2-1}{x^2+4x+2}\div\frac{x^2+2x+1}{x^2+3x+2} \vspace{5mm}\ = \frac{(x+1)(x-1)}{(x+2)(x+2)}\div\frac{(x+1)(x+1)}{(x+2)(x+1)}$

Next let's remember how we divide one fraction by another—by multiplying by the reciprocal:

$\frac{(x+1)(x-1)}{(x+2)(x+2)}\div\frac{(x+1)(x+1)}{(x+2)(x+1)} \vspace{5mm}\ = \frac{(x+1)(x-1)}{(x+2)(x+2)}\times\frac{(x+2)(x+1)}{(x+1)(x+1)}$

In this form, we can see that a lot of terms are going to start canceling with each other. All that we're left with is just $\frac{x-1}{x+2}$ .

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Question

Determine the domain of

$\frac{x+1}{x-1}$

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Answer

Because the denominator cannot be zero, the domain is all other numbers except for 1, or

$x\neq1$

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Question

Which of the following is the best definition of a rational expression?

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Answer

The rational expression is a ratio of two polynomials.

$\frac{A(x)}{B(x)}$

The denominator cannot be zero.

An example of a rational expression is:

$\frac{x^2+x-5}{x+8}$

The answer is:

$\textup{The ratio of two polynomials that cannot have a zero denominator.}$

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Question

$f(x)=\frac{2x-5}{x+3}-\frac{-3x-1}{x-1}$

Which of the following equations is equivalent to ?

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Answer

By looking at the answer choices, we can assume that the problem wants us to simplify . To do that, we need to combine the two terms within into one fraction.

First, let's remember how to add or subtract fractions:

Make sure the fractions have the same denominator.

Add or subtract the numerators, leaving the denominator alone.

The process looks like this:

$\frac{a}{b}+\frac{c}{d}=\frac{a}{b}\left( \frac{d}{d} \right) + \frac{c}{d}\left( \frac{b}{b} \right) = \frac{ad}{bd}+\frac{cb}{bd}=\frac{ad+cb}{bd}$

This is exactly what we're going to have to do to .

$f(x)=\frac{2x-5}{x+3}-\frac{-3x-1}{x-1}$

First, we find a common denominator between the two terms. No matter what ends up being equal to, a common denominator can always be found by multiplying the two terms together. In other words, we can use as our common denominator.

$f(x)=\frac{2x-5}{x+3}\left ( \frac{x-1}{x-1} \right )-\frac{-3x-1}{x-1} \left ( \frac{x+3}{x+3} \right )$

$f(x)=\frac{(2x-5)(x-1)}{(x+3)(x-1)}-\frac{(-3x-1)(x+3)}{(x-1)(x+3)}$

$f(x)=\frac{(2x-5)(x-1)-(-3x-1)(x+3)}{(x+3)(x-1)}$

Now, all that's left is getting rid of these parentheses.

$f(x)=\frac{(2x^{2}-2x-5x+5)-(-3x^{2}-9x-x-3)}{x^{2}-x+3x-3}$

$f(x)=\frac{2x^{2}-7x+5+3x^{2}+10x+3}{x^{2}+2x-3}$

$f(x)=\frac{5x^{2}+3x+8}{x^{2}+2x-3}$

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Question

$\frac{(x^{2}-1x-6)}{x+3}+\frac{(x^{2}-6x+9)}{x+2}$

What is the least common denominator of the above expression?

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Answer

The least common denominator is the least common multiple of the denominators of a set of fractions.

Simply multiply the two denominators together to find the LCD:

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Question

Simplify the expression:

$\frac{5}{x+ 6} + \frac{2}{x^{2}+ 4x-12}$

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Answer

Factor the second denominator, then simplify:

$\frac{5}{x+ 6} + \frac{2}{x^{2}+ 4x-12}$

$=\frac{5}{x+ 6} + \frac{2}{(x-2)(x+6)}$

$=\frac{5 (x-2)}{ (x-2)(x+6)} + \frac{2}{(x-2)(x+6)}$

$=\frac{5x-10}{ (x-2)(x+6)} + \frac{2}{(x-2)(x+6)}$

$=\frac{5x-10 + 2}{ (x-2)(x+6)}$

$=\frac{5x-8}{ (x-2)(x+6)}$

$=\frac{5x-8}{x^{2}+ 4x-12}$

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Question

Find the least common denominator of the following fractions:

$\frac{6}{7} , \frac{4}{3} , \frac{5}{9}$

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Answer

The denominators are 7, 3, and 9. We have to find the common multiple of 7, 3, and 9.

Multiples of 7: 7, 14, 21, 28, 35, 42, 49, 56, 63, 70

Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63

Multiples of 9: 9, 18, 27, 36, 45, 54, 63, 72, 81, 90

The least common multiple of the 3 denominators is 63.

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