Rational Expressions - ACT Math

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Question

Simplify the following rational expression:

$\frac{5x-5}{2x^{2}} + \frac{7x+9}{2x^{2}}$

Answer

Since both rational terms in the expression have the common denominator $2x^{2}$ , combine the numerators and simplify like terms:

$\frac{5x-5}{2x^{2}} + \frac{7x+9}{2x^{2}}$

$=\frac{(5x-5)+(7x+9)}{2x^{2}}$

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

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

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Question

Simplify the following rational expression: (9x - 2)/(x2) MINUS (6x - 8)/(x2)

Answer

Since both expressions have a common denominator, x2, we can just recopy the denominator and focus on the numerators. We get (9x - 2) - (6x - 8). We must distribute the negative sign over the 6x - 8 expression which gives us 9x - 2 - 6x + 8 ( -2 minus a -8 gives a +6 since a negative and negative make a positive). The numerator is therefore 3x + 6.

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Question

Simplify the following rational expression:

$\frac{7x-18}{x^{2}}+\frac{6x-14}{x^{2}}$

Answer

Since both fractions in the expression have a common denominator of $x^{2}$ , we can combine like terms into a single numerator over the denominator:

$\frac{7x-18}{x^{2}}+\frac{6x-14}{x^{2}}$

$=\frac{(7x-18)+(6x-14)}{x^{2}}$

$=\frac{13x-32}{x^{2}}$

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Question

Simplify the following expression:

$\frac{10x-9}{x^{3}}+\frac{11x+12}{x^{3}}$

Answer

Since both terms in the expression have the common denominator $x^{3}$ , combine the fractions and simplify the numerators:

$\frac{10x-9}{x^{3}}+\frac{11x+12}{x^{3}}$

$=\frac{(10x-9)+(11x+12)}{x^{3}}$

$=\frac{21x+3}{x^{3}}$

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Question

Compute the following: $\frac{x+2}{2-x}+\frac{x-2}{2-x}$

Answer

Notice that the denominator are the same for both terms. Since they are both the same, the fractions can be added. The denominator will not change in this problem.

$\frac{x+2}{2-x}+\frac{x-2}{2-x} = \frac{x+2+x-2}{2-x} = \frac{2x}{2-x}$

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Question

Combine the following rational expressions:

$\frac{3x^2-4x+5}{x+1}+\frac{4x^2-5x+9}{x+1}$

Answer

When working with complex fractions, it is important not to let them intimidate you. They follow the same rules as regular fractions!

$\frac{3x^2-4x+5}{x+1}+\frac{4x^2-5x+9}{x+1}$

In this case, our problem is made easier by the fact that we already have a common denominator. Nothing fancy is required to start. Simply add the numerators:

$\frac{(3x^2-4x+5)+(4x^2-5x+9)}{x+1}$

For our next step, we need to combine like terms. This is easier to see if we group them together.

$\frac{({\color{Blue} 3x^2+4x^2})+({\color{Red} -4x+-5x})+({\color{DarkOrange} 5+9})}{x+1}$

$\frac{({\color{Blue}7x^2})+({\color{Red} -9x})+({\color{DarkOrange} 14})}{x+1}$

Thus, our final answer is:

$\frac{7x^2 -9x+14}{x+1}$

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Question

Simplify the following expression:

$\frac{5}{7}+\frac{11}{14}$

Answer

In order to add fractions, we must first make sure they have the same denominator.

So, we multiply $\frac{5}{7}$ by $\frac{2}{2}$ and get the following:

$\frac{5}{7}\left(\frac{2}{2}\right)+\frac{11}{14}$

$\frac{10}{14}+\frac{11}{14}$

Then, we add across the numerators and simplify:

$\frac{10}{14}+\frac{11}{14}=\frac{21}{14}=\frac{3}{2}$

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Question

Simplify the following:

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

Answer

To simplify the following, a common denominator must be achieved. In this case, the first term must be multiplied by (x+2) in both the numerator and denominator and likewise with the second term with (x-3).

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

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

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Question

Simplify the following $\frac{3x}{x-3} + \frac{7}{x-4}$

Answer

Find the least common denominator between x-3 and x-4, which is (x-3)(x-4). Therefore, you have $\frac{3x(x-4)}{(x-3)(x-4)} + \frac{7(x-3)}{(x-4)(x-3)}$ . Multiplying the terms out equals $\frac{3x^{2}-12x+ 7x-21}{x^{2}-7x+12}$ . Combining like terms results in $\frac{3x^{2}-5x-21}{x^{2}-7x+12}$ .

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Question

Select the expression that is equivalent to

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

Answer

To add the two fractions, a common denominator must be found. With one-term denominators, it is easier to simply find the least common denominator between them and multiply each side to obtain it.

In this case, the least common denominator between and is . So the first fraction needs to be multiplied by and the second by :

$\frac{x^2 + 3x - 2}{3x} + \frac{x-3}{x^2} = \frac{x^2 + 3x - 2}{3x}* \left(\frac{x}{x}\right) + \frac{x-3}{x^2} * \left(\frac{3}{3}\right)$

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

Now, we can add straight across, remembering to combine terms where we can.

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

So, our simplified answer is $\frac{x^3 + 3x^2 +x -9}{3x^2}$

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Question

Combine the following two expressions if possible.

$\frac{x+3}{x+7} + \frac{x^2}{4-x}$

Answer

For binomial expressions, it is often faster to simply FOIL them together to find a common trinomial than it is to look for individual least common denominators. Let's do that here:

$\frac{x+3}{x+7} + \frac{x^2}{4-x} = \frac{x+3}{x+7} * \frac{(4-x)}{(4-x)} + \frac{x^2}{4-x} * \frac{(x+7)}{(x+7)}$

FOIL and simplify.

$\frac{-x^2 + x + 12}{-x^2 -3x+28} + \frac{x^3 +7x^2}{-x^2 -3x+28}$

Combine numerators.

$\frac{-x^2 + x + 12}{-x^2 -3x+28} + \frac{x^3 +7x^2}{-x^2 -3x+28} = \frac{x^3 +6x^2 + x + 12}{-x^2-3x+28}$

Thus, our answer is $\frac{x^3 +6x^2 + x + 12}{-x^2-3x+28}$

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Question

Simplify the expression, leaving no radicals in the denominator:

$\frac{3+\sqrt{2}}{3 -\sqrt{3}}$

Answer

The easy way to solve this problem is to multiply both halves of the fraction by the conjugate of the denominator, since this will eliminate the radical in the denominator.

$\frac{3+\sqrt{2}}{3 -\sqrt{3}} \cdot \frac{3+\sqrt{3}}{3+\sqrt{3}} = \frac{9+3\sqrt{2} + 3\sqrt{3} + \sqrt{6}}{9-3\sqrt{3}+3\sqrt{3}-3}$ Conjugate the fraction.

Next, simplify the denominator, eliminating any terms you can along the way.

$\frac{9+3\sqrt{2} + 3\sqrt{3} + \sqrt{6}}{9-3\sqrt{3}+3\sqrt{3}-3} = \frac{9+3\sqrt{2} + 3\sqrt{3} + \sqrt{6}}{6}$

Thus, $\frac{9+3\sqrt{2} + 3\sqrt{3} + \sqrt{6}}{6}$ is our answer.

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Question

Which of the following is equivalent to $\dpi{100} \frac{(\frac{1}{t}-\frac{1}{x})}{x-t}$ ? Assume that denominators are always nonzero.

Answer

We will need to simplify the expression $\frac{(\frac{1}{t}-\frac{1}{x})}{x-t}$ . We can think of this as a large fraction with a numerator of $\frac{1}{t}-\frac{1}{x}$ and a denominator of $\dpi{100} x-t$ .

In order to simplify the numerator, we will need to combine the two fractions. When adding or subtracting fractions, we must have a common denominator. $\frac{1}{t}$ has a denominator of $\dpi{100} t$ , and $\dpi{100} -\frac{1}{x}$ has a denominator of $\dpi{100} x$ . The least common denominator that these two fractions have in common is $\dpi{100} xt$ . Thus, we are going to write equivalent fractions with denominators of $\dpi{100} xt$ .

In order to convert the fraction $\dpi{100} \frac{1}{t}$ to a denominator with $\dpi{100} xt$ , we will need to multiply the top and bottom by $\dpi{100} x$ .

$\frac{1}{t}=\frac{1\cdot x}{t\cdot x}=\frac{x}{xt}$

Similarly, we will multiply the top and bottom of $\dpi{100} -\frac{1}{x}$ by $\dpi{100} t$ .

$\frac{1}{x}=\frac{1\cdot t}{x\cdot t}=\frac{t}{xt}$

We can now rewrite $\frac{1}{t}-\frac{1}{x}$ as follows:

$\frac{1}{t}-\frac{1}{x}$ = $\frac{x}{xt}-\frac{t}{xt}=\frac{x-t}{xt}$

Let's go back to the original fraction $\frac{(\frac{1}{t}-\frac{1}{x})}{x-t}$ . We will now rewrite the numerator:

$\frac{(\frac{1}{t}-\frac{1}{x})}{x-t}$ = $\frac{\frac{x-t}{xt}}{x-t}$

To simplify this further, we can think of $\frac{\frac{x-t}{xt}}{x-t}$ as the same as $\frac{x-t}{xt}\div (x-t)$ . When we divide a fraction by another quantity, this is the same as multiplying the fraction by the reciprocal of that quantity. In other words, $a\div b=a\cdot \frac{1}{b}$ .

$\frac{x-t}{xt}\div (x-t)$ = $\frac{x-t}{xt}\cdot \frac{1}{x-t}$ $=\frac{x-t}{xt(x-t)}$ $= \frac{1}{xt}$

Lastly, we will use the property of exponents which states that, in general, $\frac{1}{a}=a^{-1}$ .

$\frac{1}{xt}=(xt)^{-1}$

The answer is $(xt)^{-1}$ .

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Question

Simplify:

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

Answer

Multiply by the reciprocal of $\frac{x-1}1{}$ .

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

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

Divide by common factors.

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

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Question

What is $\frac{x^{2} + 11x + 28}{x+7}$ ?

Answer

$x^{2} + 11x + 28$ factors to . Thus, $\frac{(x+4)(x+7)}{x+7}$ . Canceling out like terms leads to .

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Question

Simplify the following expression:

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

Answer

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

There are a couple ways to go about solving this problem. One could simply take the reciprocal of the second fraction, multiply everything out, and then look for ways to simplify. However, it is almost always easier to simplify before doing any multiplying.

To begin, we need to take the reciprocal of the second fraction, so that our expression becomes:

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

Then, before we multiply anything out, try to factor out the different parts of this expression. If we have any common factors in the numerator and denominator, we can cancel them out. In this case, the above expression factors as follows:

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

Conveniently, we can cancel out a bunch of factors here!

The common factors are highlighted in corresponding colors. We can actually cancel everything out here. We have a in both the numerator and the denominator, and also an term. We also have two terms in both the numerator and the denominator, so all of these terms will cancel, and we will be left with :

$\frac{({\color{Red} x-1})({\color{Blue} x+3})}{({\color{Red} x-1})({\color{Red} x-1})}\ast \frac{{\color{Green} 3}({\color{Red} x-1})}{{\color{Green} 3}({\color{Blue} x+3})}=1$

The reason we can cancel out terms when they are in the numerator and denominator is that we are essentially multiplying and then dividing by the same term. Since multiplication and division are opposite operations, they cancel each other out and we are left with .

Important side note: you can only cancel across two different fractions when you are multiplying and/or dividing. You CANNOT cancel factors across fractions if you are adding or subtracting them.

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Question

Simplify the expression, leaving no radicals in the denominator:

$\frac{\sqrt{18}-\sqrt{5}}{3 -\sqrt{5}}$

Answer

The easy way to solve this problem is to multiply both halves of the fraction by the conjugate of the denominator, since this will eliminate the radical in the denominator.

$\frac{\sqrt{18}-\sqrt{5}}{3 -\sqrt{5}} \cdot \frac{3+\sqrt{5}}{3+\sqrt{5}} = \frac{3\sqrt{18}+\sqrt{90}-3\sqrt{5}-5}{9+3\sqrt{5}-3\sqrt{5}-5}}$ Conjugate the fraction.

Next, simplify the fraction, eliminating any terms you can along the way.

$\frac{3\sqrt{18}+\sqrt{90}-3\sqrt{5}-5}{9+3\sqrt{5}-3\sqrt{5}-5} = \frac{9\sqrt{2}+3\sqrt{10}-3\sqrt{5}-5}{4}$

Thus, $\frac{9\sqrt{2}+3\sqrt{10}-3\sqrt{5}-5}{4}$ is our answer.

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Question

Simplify the expression, leaving no radicals in the denominator:

$\frac{7-\sqrt{13}}{5 +\sqrt{13}}$

Answer

The easy way to solve this problem is to multiply both halves of the fraction by the conjugate of the denominator, since this will eliminate the radical in the denominator.

$\frac{7-\sqrt{13}}{5 +\sqrt{13}}\cdot \frac{5-\sqrt{13}}{5-\sqrt{13}} = \frac{35-7\sqrt{13}-5\sqrt{13}+13}{25-5\sqrt{13}+5\sqrt{13}-13}$ Conjugate the fraction.

Next, simplify the fraction, eliminating any terms you can along the way.

$\frac{35-7\sqrt{13}-5\sqrt{13}+13}{25-5\sqrt{13}+5\sqrt{13}-13} = \frac{48-12\sqrt{13}}{12} =4-\sqrt{13}$

Thus, $4-\sqrt{13}$ is our answer.

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Question

Simplify the expression:

$\frac{(4)\cdot(3x-24)}{(6x-48)\cdot(2x)}$

Answer

To begin, factor out the greatest common factor from each of the binomials to check for compatibility:

$\frac{(4)\cdot(3x-24)}{(6x-48)\cdot(2x)} = \frac{(4)\cdot3(x-8)}{6(x-8)\cdot(2x)}$ Factor.

Next, eliminate the common factors and simplify.

$\frac{(4)\cdot3(x-8)}{6(x-8)\cdot(2x)} = \frac{12}{12x}$ Eliminate and simplify.

Lastly, clean up the fraction.

$\frac{12}{12x} = \frac{1}{x}$

Thus, our answer is $\frac{1}{x}$ .

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Question

Simplify:

$\frac{(-5)\cdot(5x+30)}{(-x-6)\cdot(5x)}$

Answer

To begin, factor out the greatest common factor from each of the binomials to check for compatibility:

$\frac{(-5)\cdot(5x+30)}{(-x-6)\cdot(5x)} = \frac{(-5)\cdot5(x+6)}{-1(x+6)\cdot(5x)}$ Factor.

Next, eliminate the common factors and simplify.

$\frac{(-5)\cdot5(x+6)}{-1(x+6)\cdot(5x)} = \frac{-25}{-5x}$ Eliminate and simplify.

Lastly, clean up the fraction.

$\frac{-25}{-5x} = \frac{5}{x}$

Thus, our answer is $\frac{5}{x}$ .

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