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Algebra II : Intermediate Single-Variable Algebra

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #23 : Solving Rational Expressions

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

Possible Answers:

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

$-\frac{8}{x^2+2x-3}$

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

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

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

Correct answer:

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

Explanation:

Find the least common denominator by multiplying the denominators together.

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

Simplify the terms.

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

Convert both fractions.

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

Simplify the numerator and denominator.

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

Combine to form one fraction. Brace the numerator of the second term since we are subtracting a quantity.

$\frac{6+2x-[2-2x]}{(1-x)(3+x)} =\frac{6+2x-2+2x}{(1-x)(3+x)} =\frac{4x+4}{(1-x)(3+x)}$

Replace the denominator with the simplified form.

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

Pull out a common factor of negative one from the denominator.

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

This allows us to pull the negative sign out in front of the fraction.

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

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Example Question #29 : Adding And Subtracting Rational Expressions

Add the fractions: $\frac{2}{x}+\frac{2}{5x^2+3}$

Possible Answers:

$\frac{10x+8}{5x^2+3}$

$\frac{12x^2+6}{5x^3+3x}$

$\frac{10x^2+2x+6}{5x^3+3x}$

$\frac{10x^2+8}{5x^3+3x}$

$\frac{10x^2+2x+8}{5x^3+3x}$

Correct answer:

$\frac{10x^2+2x+6}{5x^3+3x}$

Explanation:

Determine the least common denominator by multiplying both denominator together. Convert the fractions.

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

Combine the fractions as one fractions and simplify the numerator.

$\frac{2(5x^2+3)+2x}{x(5x^2+3)}= \frac{10x^2+6+2x}{x(5x^2+3)}= \frac{10x^2+2x+6}{5x^3+3x}$

The answer is: $\frac{10x^2+2x+6}{5x^3+3x}$

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Example Question #24 : Solving Rational Expressions

Solve: $\frac{6}{2+x} -\frac{11}{x}$

Possible Answers:

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

$\frac{17x+22}{x^2+2}$

$-\frac{17x+22}{x^2+2}$

$\frac{17x-22}{x^2+2}$

$-\frac{5x+22}{x^2+2}$

Correct answer:

$-\frac{5x+22}{x^2+2}$

Explanation:

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

Multiply the two denominators together.

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

Convert both fractions to have the same common denominator.

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

Simplify the second term.

$\frac{11(2+x)}{x^2+2} = \frac{22+11x}{x^2+2}$

Combine the fractions to one fraction. Use a parentheses to brace (22+11x) term in the numerator since we are subtracting a quantity.

$\frac{6x-(22+11x)}{x^2+2} = \frac{6x-22-11x}{x^2+2}$

Combine like-terms.

$\frac{-5x-22}{x^2+2}$

Factor out a negative one from the numerator.

$\frac{-1(5x+22)}{x^2+2}$

This allows us to pull the negative sign out in front of the fraction.

The answer is: $-\frac{5x+22}{x^2+2}$

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Example Question #1731 : Algebra Ii

Add: $\frac{2}{9x}+\frac{4}{27}$

Possible Answers:

$\frac{4x+6}{27x}$

$\frac{6x+4}{27x}$

$\frac{27}{4}+ \frac{2}{9}x$

$\frac{27}{4}+ \frac{2}{9x}$

$\frac{6x+4}{27}$

Correct answer:

$\frac{4x+6}{27x}$

Explanation:

Find the least common denominator to solve this problem

Multiply 27 with , and multiply with 3 to obtain common denominators.

Convert the fractions.

$\frac{2}{9x}+\frac{4}{27}=\frac{6}{27x}+\frac{4x}{27x}$

Combine the terms as one fraction.

The answer is: $\frac{4x+6}{27x}$

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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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Algebra II : Intermediate Single-Variable Algebra

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #23 : Solving Rational Expressions

Example Question #29 : Adding And Subtracting Rational Expressions

Example Question #24 : Solving Rational Expressions

Example Question #1731 : Algebra Ii

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

All Algebra II Resources

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