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Algebra II : Algebra II

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Example Questions

Example Question #531 : Intermediate Single Variable Algebra

Find the values of which will make this rational expression undefined:

$\frac{x-1}{(x+2)(x-10)}$

Possible Answers:

x=-10,2

x=-2,1

x=-2,10

x=-1,2

x=-2,1,10

Correct answer:

x=-2,10

Explanation:

For a rational expression to be undefined, the denominator must be equal to .

1. Set the denominator equal to .

(x+2)(x-10)=0

2. Set the factors equal to and solve for .

x+2=0

x=-2

and

x-10=0

x=10

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Example Question #532 : Intermediate Single Variable Algebra

Which value of makes the following expression undefined?

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

Possible Answers:

$\text{All real numbers}$

$\text{Cannot be determined}$

Correct answer:

Explanation:

A rational expression is undefined when the denominator is zero.

x-1=0

x=1

The denominator is zero when x=1 .

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Example Question #2 : Properties Of Fractions

Compute: $\frac{x}{z}+\frac{y}{x}+\frac{z}{y}$

Possible Answers:

xyz

$\frac{x^2+y^2+z^2}{xyz}$

$\frac{x^2y+y^2z+xz^2}{xyz}$

$\frac{1}{xyz}$

Correct answer:

$\frac{x^2y+y^2z+xz^2}{xyz}$

Explanation:

The terms cannot be added unless the denominators are common. To get the same denominator for each term, find the LCD by multiplying , , and .

The denominator for the three terms are xyz . Evaluate each term of the given problem.

$\frac{x}{z}+\frac{y}{x}+\frac{z}{y}=\frac{x^2y}{xyz}+\frac{y^2z}{xyz}+\frac{xz^2}{xyz}=\frac{x^2y+y^2z+xz^2}{xyz}$

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Example Question #4 : Properties Of Fractions

Bob is left with a whole pizza for dinner. He eats $\frac{1}{3}$ of this pizza. Afterward, Wendy eats the rest of the pizza except the crusts, and leaves $\frac{1}{20}$ of the whole pizza remaining. What fraction did Wendy eat?

Possible Answers:

$\frac{37}{40}$

$\frac{59}{60}$

$\frac{13}{20}$

$\frac{3}{40}$

$\frac{37}{60}$

Correct answer:

$\frac{37}{60}$

Explanation:

A whole pizza is one unit.

After Bob ate $\frac{1}{3}$ of this pizza, $\frac{2}{3}$ will remain.

Wendy eats an amount of the two-thirds of the pizza, and leaves $\frac{1}{20}$ of the whole pizza. Write the equation and solve for .

$\frac{2}{3}-x=\frac{1}{20}$

$\frac{40}{60}-x=\frac{3}{60}$

$x=\frac{37}{60}$

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Example Question #3 : Properties Of Fractions

Simplify, if possible: $\frac{1}{a}-\frac{2a+2b}{a-b}$

Possible Answers:

$\frac{-4a^2-2a+2ab}{a^2-ab}$

$\frac{a-b-2a^2+2ab}{a^2-ab}$

$\frac{1-2a-2b}{b}$

$\frac{a-b-2a^2-2ab}{a^2-ab}$

$\frac{1}{a-b}$

Correct answer:

$\frac{a-b-2a^2-2ab}{a^2-ab}$

Explanation:

In order to subtract the numerators, first find the least common denominator in the problem. This can usually be found by multiplying the unlike denominators together.

a(a-b)=a^2-ab

For each term, multiply the numerator with what was multiplied on the denominator to get the least common denominator.

$\frac{1(a-b)}{a(a-b)}-\frac{(2a+2b)(a)}{(a-b)(a)}$

Simplify.

$=\frac{a-b}{a^2-ab}-\frac{2a^2+2ab}{a^2-ab}$

$= \frac{a-b-(2a^2+2ab)}{a^2-ab}$

$= \frac{a-b-2a^2-2ab}{a^2-ab}$

There are no other terms or common factors here that can be simplified. Since there are no like terms in the numerator or denominator, this is fully simplified.

The answer is:

$\frac{a-b-2a^2-2ab}{a^2-ab}$

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Example Question #4 : Properties Of Fractions

Simplify: $\frac{1}{3x}-\frac{1}{3-x}$

Possible Answers:

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

$\frac{x}{9-x}$

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

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

$-\frac{4x}{9}$

Correct answer:

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

Explanation:

In order to simplify, we will need a common denominator.

Multiply the denominators together.

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

This will be the new denominator.

Convert the fractions.

$\frac{1}{3x}-\frac{1}{3-x}=\frac{3-x}{ 9x-3x^2}-\frac{3x}{ 9x-3x^2}$

Combine the fractions as a single fraction.

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

Subtract the numerators.

The answer is: $\frac{3-4x}{ 9x-3x^2}$

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Example Question #11 : Rational Expressions

Add the fractions: $\frac{1}{3}+\frac{2}{4x^2}$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

To add these fractions, we must have a common denominator.

Multiply both denominators to obtain the least common denominator.

3(4x^2)=12x^2

Convert both fractions.

$\frac{1}{3}+\frac{2}{4x^2}=\frac{4x^2}{12x^2}+\frac{2(3)}{12x^2}$

Simplify the numerator and combine the fractions as a single fraction.

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

Pull out a common factor on the numerator.

$\frac{4x^2+6}{12x^2}=\frac{2(2x^2+3)}{12x^2}$

Reduce the fraction.

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

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Example Question #12 : Rational Expressions

Which of the following is a misuse of fractional properties?

Possible Answers:

$\frac{a+b}{b+c}\div\frac{a+c}{b+d} = \frac{ab+ad+b^2+bd}{ab+ac+bc+c^2}$

$\frac{x}{y}\times \frac{w}{z} = \frac{xw}{yz}$

$\textup{All of the given scenarios are correct.}$

$\frac{b}{a}-\frac{d}{c} = \frac{bc}{ac}-\frac{ad}{ac}$

$\frac{a}{c}+\frac{b}{d}+\frac{c}{e} =\frac{ac+bd+ce}{cde}$

Correct answer:

$\frac{a}{c}+\frac{b}{d}+\frac{c}{e} =\frac{ac+bd+ce}{cde}$

Explanation:

For the additional and subtraction properties of fractions, we must determine the least common factor, change the fractions, and find the value of the numerators.

$\frac{a}{c}+\frac{b}{d}+\frac{c}{e} = \frac{a(de)}{c(de)}+\frac{b(ce)}{d(ce)}+\frac{c(cd)}{e(cd)}$

Combining all the terms, this fraction will become:

$\frac{a}{c}+\frac{b}{d}+\frac{c}{e} =\frac{ade+bce+c^2d}{cde}$

According to the multiplicative rule of fractions, we can simply multiply the numerators together, and denominators together.

For division, the division sign can be switched to a multiplication sign, but we must also take the reciprocal of what is being divided.

We can verify that:

$\frac{a+b}{b+c}\div\frac{a+c}{b+d} = \frac{ab+ad+b^2+bd}{ab+ac+bc+c^2}$

$\frac{a+b}{b+c}\div\frac{a+c}{b+d}=\frac{a+b}{b+c}\times \frac{b+d}{a+c} = \frac{(a+b)(b+d)}{(b+c)(a+c)}$

(a+b)(b+d) = ab+ad+b^2+bd

(b+c)(a+c) = ba+bc+ca+c^2

Both terms in the numerator and denominator are satisfied by FOIL method.

The answer is: $\frac{a}{c}+\frac{b}{d}+\frac{c}{e} =\frac{ac+bd+ce}{cde}$

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Example Question #13 : Rational Expressions

Simply the expression:

$\frac{3}{x+5}+ \frac{7}{x^{2}+3x -10}$

Possible Answers:

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

$\frac{7x+5}{x^{2}+4x-5}$

$\frac{3x+1}{x^{2}+3x-10}$

$\frac{7x-3}{x^{2}+3x-10}$

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

Correct answer:

$\frac{3x+1}{x^{2}+3x-10}$

Explanation:

In order to simplify the expression $\frac{3}{x+5}+ \frac{7}{x^{2}+3x -10}$ , first note that the denominators in both terms share a factor:

$\frac{3}{x+5}+ \frac{7}{(x+5)(x-2)}$

Find the Least Common Denominator (LCD) of both terms, and then simplify the expression:

$\frac{3(x-2)}{(x+5)(x-2)}+ \frac{7}{(x+5)(x-2)}$

$=\frac{3x-6}{(x+5)(x-2)}+ \frac{7}{(x+5)(x-2)}$

$=\frac{3x-6+7}{(x+5)(x-2)}$

Which equals:

$\frac{3x+1}{x^{2}+3x-10}$

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Example Question #1 : Least Common Denominator

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

Which of the following equations is equivalent to f(x) ?

Possible Answers:

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

$f(x)=\frac{4}{5x-4}$

$f(x)=\frac{5x-4}{4}$

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

Correct answer:

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

Explanation:

By looking at the answer choices, we can assume that the problem wants us to simplify f(x) . To do that, we need to combine the two terms within f(x) 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) .

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

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 (x+3)(x-1) 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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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #531 : Intermediate Single Variable Algebra

Example Question #532 : Intermediate Single Variable Algebra

Example Question #2 : Properties Of Fractions

Example Question #4 : Properties Of Fractions

Example Question #3 : Properties Of Fractions

Example Question #4 : Properties Of Fractions

Example Question #11 : Rational Expressions

Example Question #12 : Rational Expressions

Example Question #13 : Rational Expressions

Example Question #1 : Least Common Denominator

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