Properties of Fractions - Algebra II

Card 0 of 13

Question

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

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

Answer

If $\left ( x-1 \right ) = 0$ or $\left ( x+2 \right ) =0$ , the denominator is 0, which makes the expression undefined.

This happens when x = 1 or when x = -2.

Therefore the correct answer is .

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Question

Simply the expression:

$\frac{4}{x+6}+ \frac{9}{x+4}$

Answer

In order to simplify the expression $\frac{4}{x+6}+ \frac{9}{x+4}$ , we need to ensure that both terms have the same denominator. In order to do so, find the Least Common Denominator (LCD) for both terms and simplify the expression accordingly:

$\frac{4(x+4)}{(x+6)(x+4)} + \frac{9(x+6)}{(x+4)(x+6)}$

$=\frac{4(x+4) +9(x+6)}{(x+6)(x+4)}$

$=\frac{4x+16+9x+54}{x^{2}+10x+24}$

$=\frac{13x+70}{x^{2}+10x+24}$

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Question

Simply the expression:

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

Answer

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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Question

Simplify the expression:

$\frac{x-7}{x+3}+ \frac{9}{x^{2}+10x+21}$

Answer

In order to simplify the expression $\frac{x-7}{x+3}+ \frac{9}{x^{2}+10x+21}$ , first note that the denominators in both terms share a factor:

$\frac{x-7}{x+3}+ \frac{9}{(x+3)(x+7)}$

Find the Least Common Denominator (LCD) of both terms:

$\frac{(x-7)(x+7)}{(x+3)(x+7)}+ \frac{9}{(x+3)(x+7)}$

$=\frac{x^{2}-49}{x^{2}+10x+21}+ \frac{9}{x^{2}+10x+21}$

Finally, combine like terms:

$\frac{x^{2}-49+9}{x^{2}+10x+21}$

$=\frac{x^{2}-40}{x^{2}+10x+21}$

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Question

Which value of makes the following expression undefined?

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

Answer

A rational expression is undefined when the denominator is zero.

The denominator is zero when .

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Question

Simplify the expression:

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

Answer

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

1. Create a common denominator between the two fractions.

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

2. Simplify.

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

$=\frac{5x+15-10x-20}{x^{2}+5x+6}$

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

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Question

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

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

Answer

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

1. Set the denominator equal to .

2. Set the factors equal to and solve for .

and

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Question

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

Answer

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 . 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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Question

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?

Answer

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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Question

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

Answer

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.

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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Question

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

Answer

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

Multiply the denominators together.

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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Question

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

Answer

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

Multiply both denominators to obtain the least common denominator.

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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Question

Which of the following is a misuse of fractional properties?

Answer

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:

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

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

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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