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ACT Math : Expressions

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Example Question #1 : How To Subtract Rational Expressions With Different Denominators

Simplify:

$\frac{4x}{(x+4)}-\frac{2}{x^2+6x+8}$

Possible Answers:

$\frac{4x^2+6x-8}{x^2+6x+8}$

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

$\frac{4x^2+8x-2}{x^2+6x+8}$

$\frac{4x^2+6x-1}{x^2+6x+8}$

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

Correct answer:

$\frac{4x^2+8x-2}{x^2+6x+8}$

Explanation:

Always start the simplification of rational expressions with quadratic denominators by factoring the denominator that has a squared element:

$\frac{4x}{(x+4)}-\frac{2}{x^2+6x+8} = \frac{4x}{(x+4)}-\frac{2}{(x+4)(x+2)}$

Thus, we need to multiply the numerator and denominator of the first term by (x+2) :

$\frac{4x}{(x+4)}*\frac{x+2}{x+2}-\frac{2}{(x+4)(x+2)}$

Distribute:

$\frac{4x^2+8x}{(x+4)(x+2)}-\frac{2}{(x+4)(x+2)}$

This is simply:

$\frac{4x^2+8x-2}{(x+4)(x+2)}$

You cannot factor the top in any way that would help you to cancel anything. Therefore, the answer stands as it does. You need to FOIL the denominator back out, as your correct answer is in that form:

$\frac{4x^2+8x-2}{x^2+6x+8}$

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Example Question #2 : How To Subtract Rational Expressions With Different Denominators

Simplify:

$\frac{4x}{x^2-x-6}-\frac{2x-4}{x^2-9}$

Possible Answers:

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

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

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

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

$\frac{2x^2+12x+8}{(x-3)(x+3)(x+2)}$

Correct answer:

$\frac{2x^2+12x+8}{(x-3)(x+3)(x+2)}$

Explanation:

Begin by factoring your denominators:

$\frac{4x}{x^2-x-6}-\frac{2x-4}{x^2-9}=\frac{4x}{(x-3)(x+2)}-\frac{2x-4}{(x+3)(x-3)}$

Therefore, the common denominator of our fractions is:

(x-3)(x+3)(x+2)

This will require you to multiply the first fraction by $\frac{x+3}{x+3}$ and the second by $\frac{x+2}{x+2}$ . This gives you:

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

Now, FOIL the second member and distribute the first:

$\frac{4x^2+12x-(2x^2-8)}{(x-3)(x+3)(x+2)}$

Next, distribute the subtraction:

$\frac{4x^2+12x-2x^2+8}{(x-3)(x+3)(x+2)}$

Finally, simplify:

$\frac{2x^2+12x+8}{(x-3)(x+3)(x+2)}$

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Example Question #11 : Rational Expressions

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

Possible Answers:

x +4

x + 11

x -4

$x^{2} +4$

x+7

Correct answer:

x +4

Explanation:

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

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Example Question #1 : Expressions

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

Possible Answers:

$x-t$

$\frac{x}{t}$

$(xt)^{-1}$

$x^{2}-t^{2}$

$t-x$

Correct answer:

$(xt)^{-1}$

Explanation:

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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Example Question #1 : How To Divide Rational Expressions

Simplify:

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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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Example Question #12 : Rational Expressions

Simplify the following expression:

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

Possible Answers:

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

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

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

Correct answer:

Explanation:

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

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 (x+3) term. We also have two (x-1) 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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Example Question #2411 : Act Math

Simplify the expression, leaving no radicals in the denominator:

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

Possible Answers:

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

$\frac{9+3\sqrt{2} + 3\sqrt{3}}{2}$

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

$\frac{9+3\sqrt{5} + \sqrt{6}}{6}$

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

Correct answer:

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

Explanation:

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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Example Question #14 : Rational Expressions

Simplify the expression, leaving no radicals in the denominator:

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

Possible Answers:

$4-\sqrt{13}$

$\frac{25-12\sqrt{13}}{12}$

$\frac{48-13\sqrt{12}}{13}$

$\frac{48-12\sqrt{13}}{13}$

$4+\sqrt{13}$

Correct answer:

$4-\sqrt{13}$

Explanation:

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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Example Question #15 : Rational Expressions

Simplify the expression, leaving no radicals in the denominator:

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

Possible Answers:

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

$\frac{9\sqrt{2}+3\sqrt{2}-5}{4}$

$\frac{9\sqrt{2}+3\sqrt{5}-2\sqrt{3}-5}{2}$

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

$\frac{3\sqrt{6}+3\sqrt{10}+3\sqrt{4}-5}{3}$

Correct answer:

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

Explanation:

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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Example Question #16 : Rational Expressions

Simplify the expression:

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

Possible Answers:

$-\frac{1}{x}$

$\frac{2x}{3}$

$\frac{3}{2x}$

$\frac{1}{x}$

$\frac{1}{8x}$

Correct answer:

$\frac{1}{x}$

Explanation:

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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ACT Math : Expressions

Study concepts, example questions & explanations for ACT Math

All ACT Math Resources

Example Questions

Example Question #1 : How To Subtract Rational Expressions With Different Denominators

Example Question #2 : How To Subtract Rational Expressions With Different Denominators

Example Question #11 : Rational Expressions

Example Question #1 : Expressions

Example Question #1 : How To Divide Rational Expressions

Example Question #12 : Rational Expressions

Example Question #2411 : Act Math

Example Question #14 : Rational Expressions

Example Question #15 : Rational Expressions

Example Question #16 : Rational Expressions

All ACT Math Resources

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