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Algebra II : Rational Expressions

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #1732 : Algebra Ii

Add: $\frac{x^2+2x+3}{x}+\frac{x}{x^2+2x+3}$

Possible Answers:

$\frac{x^4+45}{2x^3+4x^2+6x}$

$\frac{2x^2+4x+18}{x^3+2x^2+3x}$

$\frac{x^4+27x^3+15x^2+24x+9}{x^3+2x^2+3x}$

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

$\frac{x^4+4x^3+11x^2+12x+9}{x^3+2x^2+3x}$

Correct answer:

$\frac{x^4+4x^3+11x^2+12x+9}{x^3+2x^2+3x}$

Explanation:

In order to add the numerators of the fractions, the least common denominator is necessary.

Multiply the denominators together to get the LCD.

x(x^2+2x+3) = x^3+2x^2+3x

Convert the fractions by multiplying the top by what was multiplied on the bottom to get the LCD.

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

Factor the trinomials on the numerator of the first term.

(x^2+2x+3)(x^2+2x+3)

Multiply the first, second, and third term to the quantity of the second trinomial and sum the quantities.

x^2(x^2+2x+3)+2x(x^2+2x+3)+3(x^2+2x+3)

(x^4+2x^3+3x^2)+(2x^3+4x^2+6x)+(3x^2+6x+9)

Combine like terms.

x^4+4x^3+10x^2+12x+9

The fractions become:

$\frac{x^4+4x^3+10x^2+12x+9}{x(x^2+2x+3)}+\frac{x^2}{(x)(x^2+2x+3)}$

Combine like terms and in one fraction.

The answer is: $\frac{x^4+4x^3+11x^2+12x+9}{x^3+2x^2+3x}$

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Example Question #71 : Rational Expressions

Add: $\frac{x}{2x-1}+\frac{2x}{3x-1}$

Possible Answers:

$\frac{2x^2-3x}{6x^2-5x+1}$

$\frac{4x+2}{5x^2-1}$

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

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

$\frac{7x^2-3x}{6x^2-5x+1}$

Correct answer:

$\frac{7x^2-3x}{6x^2-5x+1}$

Explanation:

In order to add the fractions, we will need a least common denominator.

Multiply both denominators together by using the FOIL method.

(2x-1)(3x-1)

=(2x)(3x)+(2x)(-1)+(-1)(3x)+(-1)(-1)

=6x^2-2x-3x+1

=6x^2-5x+1

This will be the least common denominator. Convert the fractions. Remember to multiply the separate numerators by what was multiplied on the denominator to get the LCD.

$\frac{x}{2x-1}+\frac{2x}{3x-1} = \frac{x(3x-1)}{6x^2-5x+1}+\frac{2x(2x-1)}{6x^2-5x+1}$

Simplify the numerators by distribution.

$\frac{x(3x-1)}{6x^2-5x+1}+\frac{2x(2x-1)}{6x^2-5x+1}$

$=\frac{3x^2-x}{6x^2-5x+1}+\frac{4x^2-2x}{6x^2-5x+1}$

Combine the fractions as one whole and combine like-terms.

The answer is: $\frac{7x^2-3x}{6x^2-5x+1}$

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Example Question #72 : Rational Expressions

Solve the expression: $\frac{3x}{x^2+1}+\frac{1}{x}$

Possible Answers:

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

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

$\frac{4x+1}{ x^3+x}$

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

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

Correct answer:

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

Explanation:

To be able to add the numerators, we will need to find the least common denominator by multiplying the denominators.

x(x^2+1) = x^3+x

Convert the fractions with this denominator.

$\frac{3x}{x^2+1}+\frac{1}{x}=\frac{3x(x)}{x(x^2+1)}+\frac{1(x^2+1)}{x(x^2+1)}$

Simplify the numerators and combine as one fraction.

$\frac{3x^2}{x(x^2+1)}+\frac{x^2+1}{x(x^2+1)}=\frac{4x^2+1}{x(x^2+1)}$

The answer is: $\frac{4x^2+1}{ x^3+x}$

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Example Question #72 : Rational Expressions

Add the following terms: $\frac{x}{x+3}+\frac{x}{2x+3}$

Possible Answers:

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

$\frac{4x+3}{2x^2+9}$

$\frac{3x+6}{2x^2+8x+9}$

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

$\frac{3x^2+6x}{2x^2+9x+9}$

Correct answer:

$\frac{3x^2+6x}{2x^2+9x+9}$

Explanation:

In order to add the rational expressions, we will need a least common denominator.

Use the FOIL method to expand the binomials.

(x+3)(2x+3)= (x)(2x)+(x)(3)+(3)(2x)+(3)(3)

Simplify the terms.

= 2x^2+3x+6x+9 = 2x^2+9x+9

This is the least common denominator. Convert both fractions.

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

Simplify the numerators.

$\frac{x(2x+3)}{(x+3)(2x+3)}+\frac{x(x+3)}{(2x+3)(x+3)}$

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

Combine like terms and as one fraction.

The answer is: $\frac{3x^2+6x}{2x^2+9x+9}$

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Example Question #73 : Rational Expressions

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

Possible Answers:

$\frac{2}{x-1}$

$\frac{2x+1}{x^2-1}$

$\frac{2}{x^2-1}$

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

$-\frac{2}{x^2-1}$

Correct answer:

$\frac{2}{x^2-1}$

Explanation:

To be able to subtract the expressions, we will need to change the denominators to a least common denominator.

Multiply both denominators by FOIL method.

(x+1)(x-1)=x^2-1

Convert the fractions with the LCD.

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

The numerator of the second term can also be simplified by the FOIL method.

(x-2)(x+1) = x^2-x-2

Rewrite the fractions and combine as one fraction. Since we are subtracting a quantity, it is necessary to enclose the second numerator with parentheses.

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

Simplify the numerator.

x^2-x-(x^2-x-2) = x^2-x-x^2+x+2 = 2

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

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Example Question #31 : Adding And Subtracting Rational Expressions

Add: $\frac{8}{x}-\frac{8}{x^3-1}$

Possible Answers:

$\frac{8x-8}{x^2+x+1}$

$\frac{8x^3-x-8}{x^4-x}$

$\frac{8x^3-8x-8}{x^4-x}$

$\frac{1}{x^4-x}$

$\frac{2}{x^2-1}$

Correct answer:

$\frac{8x^3-8x-8}{x^4-x}$

Explanation:

Multiply the two denominators together to determine the least common denominator.

x(x^3-1) = x^4-x

Convert both fractions to the same denominator.

$\frac{8(x^3-1)}{x(x^3-1)}-\frac{8x}{x(x^3-1)}$

Simplify the numerator and combine as one fraction.

$\frac{8x^3-8-8x}{x(x^3-1)} = \frac{8x^3-8x-8}{x^4-x}$

The answer is: $\frac{8x^3-8x-8}{x^4-x}$

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Example Question #1738 : Algebra Ii

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

First, identify the common denominator. In this case, it's x^2 .

Offset the first fraction to get the new denominator:

$\frac{4}{x}=\frac{4x}{x^2}$

Now, combine the numerators of both fractions:

4x+3-x=3x+3

Put that over your denominator to get your answer:

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

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Example Question #601 : Intermediate Single Variable Algebra

Add: $\frac{x+9}{x-9} +\frac{3}{x^2}$

Possible Answers:

$\frac{3x-5}{3x}$

$\frac{x^3-9x^2-3x+27}{x^3-9x}$

$\frac{x^3+23x-27}{x^3-9x}$

$\frac{x^3-12x-27}{x^3-9x}$

$\frac{x^3+9x^2+3x-27}{x^3-9x}$

Correct answer:

$\frac{x^3+9x^2+3x-27}{x^3-9x}$

Explanation:

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

Multiply the uncommon denominators together.

x^2(x-9) =x^3-9x^2

Convert the fractions.

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

Simplify the terms on the numerator.

$\frac{x^3+9x^2}{x^3-9x} +\frac{3x-27}{x^3-9x}$

Combine as one fraction.

The answer is: $\frac{x^3+9x^2+3x-27}{x^3-9x}$

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Example Question #71 : Rational Expressions

Subtract: $\frac{1}{x}- \frac{x^5-x^2-1}{x^2}$

Possible Answers:

$-\frac{x^5-x^2-x-1}{x^2}$

$-\frac{x^4-x-1}{x}$

$-\frac{x^3-x-2}{x}$

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

$-\frac{x^5-x^2+x+1}{x}$

Correct answer:

$-\frac{x^5-x^2-x-1}{x^2}$

Explanation:

Solve by first finding the least common denominator. The LCD can be visually seen as x^2 since both terms are divisible by this value.

Convert the first fraction.

$\frac{1}{x}- \frac{x^5-x^2-1}{x^2} =\frac{x}{x^2}- \frac{x^5-x^2-1}{x^2}$

Combine the two fractions as one. Use parentheses to brace the second numerator.

$\frac{x-(x^5-x^2-1)}{x^2}$

Simplify the numerator. Double negatives will produce a positive.

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

Reorganize the terms from highest to lowest powers.

$\frac{-x^5+x^2+x+1}{x^2}$

Pull out a common factor of -1 so that the negative sign can be in front of the fraction.

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

The answer is: $-\frac{x^5-x^2-x-1}{x^2}$

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Example Question #1742 : Algebra Ii

Subtract: $\frac{3}{8x}-\frac{6}{16-x}$

Possible Answers:

$\frac{51x-48}{8x^2+128}$

$\frac{12x-6}{8x+9}$

$\frac{51x+48}{8x^2+128}$

$\frac{51x+45}{8x^2+128}$

$\frac{51x-45}{8x^2+128}$

Correct answer:

$\frac{51x-48}{8x^2+128}$

Explanation:

In order to simplify this rational expression, we will need to determine the least common denominator.

8x(16-x) = 128x-8x^2

Convert the fractions.

$\frac{3(16-x)}{8x(16-x)}-\frac{6(8x)}{(16-x)(8x)}$

Simplify the fractions.

$\frac{48-3x}{ 128x-8x^2}-\frac{48x}{ 128x-8x^2}$

Simplify by combining like terms. Combine as one fraction.

$\frac{48-51x}{ 128x-8x^2}$

To rewrite this in the order of powers, we can factor a negative one on the top and bottom.

$\frac{-(-48+51x)}{- (128x+8x^2)} = \frac{51x-48}{8x^2+128}$

The answer is: $\frac{51x-48}{8x^2+128}$

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Algebra II : Rational Expressions

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #1732 : Algebra Ii

Example Question #71 : Rational Expressions

Example Question #72 : Rational Expressions

Example Question #72 : Rational Expressions

Example Question #73 : Rational Expressions

Example Question #31 : Adding And Subtracting Rational Expressions

Example Question #1738 : Algebra Ii

Example Question #601 : Intermediate Single Variable Algebra

Example Question #71 : Rational Expressions

Example Question #1742 : Algebra Ii

All Algebra II Resources

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