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ACT Math : Algebra

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Example Questions

Example Question #1 : How To Divide Rational Expressions

Simplify the expression, leaving no radicals in the denominator:

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

Possible Answers:

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

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

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

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

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

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 #3 : How To Divide Rational Expressions

Simplify the expression, leaving no radicals in the denominator:

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

Possible Answers:

$4-\sqrt{13}$

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

$4+\sqrt{13}$

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

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

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 #2413 : Act Math

Simplify the expression, leaving no radicals in the denominator:

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

Possible Answers:

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

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

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

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

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

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 #2414 : Act Math

Simplify the expression:

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

Possible Answers:

$\frac{1}{x}$

$\frac{1}{8x}$

$-\frac{1}{x}$

$\frac{2x}{3}$

$\frac{3}{2x}$

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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Example Question #2 : How To Divide Rational Expressions

Simplify:

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

Possible Answers:

$-\frac{5}{x}$

$\frac{25}{x+5}$

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

$\frac{5}{x}$

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

Correct answer:

$\frac{5}{x}$

Explanation:

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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Example Question #3 : How To Divide Rational Expressions

Simplify the expression:

$\frac{(3ab-9ax)}{(9a^2b-27a^2x)}$

Possible Answers:

$\frac{1}{3a}$

$\frac{9a^2}{3a}$

$\textup{The expression is prime, and cannot be simplified.}$

$\frac{3a}{9a^2}$

Correct answer:

$\frac{1}{3a}$

Explanation:

This problem looks intimidating, but can be solved with our standard tool; factoring the GCF out of each binomial.

$\frac{(3ab-9ax)}{(9a^2b-27a^2x)} = \frac{3a(b-3x)}{9a^2(b-3x)}$

Eliminate the obvious binomial:

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

But we're not through yet! Notice that we can further simplify this fraction:

$\frac{(3a)}{(9a^2)} = \frac{3a}{3a\cdot 3a}$

Eliminate another pair, and the solution is now obvious:

$\frac{3a}{3a\cdot 3a} = \frac{1}{3a}$

Thus, our answer is $\frac{1}{3a}$ .

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Example Question #2411 : Act Math

Multiply $\left ( 3x-1\right )\cdot \left ( x+4\right )$

Possible Answers:

$3x^{2}+13x+4$

$3x^{2}+11x-4$

$3x^{2}-11x-4$

$3x^{2}+13x-4$

$3x^{2}+11x+4$

Correct answer:

$3x^{2}+11x-4$

Explanation:

When multiplying two binomials, it's best to remember the mnemonic FOIL, which stands for First, Outer, Inner, and Last. Going in the order of FOIL, we multiply the first term of each binomial, then the outer terms of the expression, then the inner terms of the expression, and finally the last term of each binomial. For this question we get: $F : 3x\cdot x = 3x^{2}, O : 3x\cdot 4 = 12x, I : -1\cdot x = -x, L: -1\cdot 4 = -4$ Now that we have 'FOILed' our expression, we simply add our results to get the final answer. $3x^{2} + 12x - x -4 = 3x^{2}+11x-4$

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Example Question #2 : How To Multiply Rational Expressions

Give the product in simplified form: $\frac{9x}{y^{2}} \cdot \frac{x^{3}}{12} \cdot \frac{2}{xy}$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Use cross-cancellation as follows:

$\frac{9x}{y^{2}} \cdot \frac{x^{3}}{12} \cdot \frac{2}{xy}$

$= \frac{3x}{y^{2}} \cdot \frac{x^{2}}{2} \cdot \frac{1}{ y}$

$= \frac{3x^{3}}{2y^{3}}$

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Example Question #2 : How To Multiply Rational Expressions

Give the product in simplified form:

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

Possible Answers:

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

$\frac{ 4y - 28 }{ 7x+ 28 }$

$\frac{ 4y - 1 }{ 7x+ 1 }$

$\frac{ y - 4 }{ x+ 7 }$

$\frac{ 4y }{ 7x }$

Correct answer:

$\frac{ 4y - 28 }{ 7x+ 28 }$

Explanation:

Factor the numerator and denominator in the second fraction, cross-cancel common factors, then multiply out.

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

$= \frac{4x}{7y} \cdot \frac{y \left (y - 7 \right ) }{x\left ( x+ 4 \right )}$

$= \frac{4 }{7 } \cdot \frac{ y - 7 }{ x+ 4 }$

$= \frac{ 4y - 28 }{ 7x+ 28 }$

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Example Question #23 : Rational Expressions

Give the product in simplified form:

$\left (x^{2} - 9 \right ) \cdot \frac{x^{2}+6x+9}{x^{2}-6x+9}$

Possible Answers:

$\frac{ x^{3} +27}{x-3}$

$\frac{ x^{3} + 9 x^{2} +27 x +27}{x-3}$

$x^{2}+6x+9$

$x^{2} + 3x+9$

$\frac{ x^{3} + 9 x^{2} +9x +27}{x-3}$

Correct answer:

$\frac{ x^{3} + 9 x^{2} +27 x +27}{x-3}$

Explanation:

Factor the first polynomial and the numerator and denominator in the second fraction, cross-cancel common factors, then multiply out.

$\left (x^{2} - 9 \right ) \cdot \frac{x^{2}+6x+9}{x^{2}-6x+9}$

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

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

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

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

x+3 can be cubed using the cube of a binomial pattern:

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

$= x^{3} + 9 x^{2} +27 x +27$

so the above rational expression can be rewritten as

$\frac{ x^{3} + 9 x^{2} +27 x +27}{x-3}$

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ACT Math : Algebra

Study concepts, example questions & explanations for ACT Math

All ACT Math Resources

Example Questions

Example Question #1 : How To Divide Rational Expressions

Example Question #3 : How To Divide Rational Expressions

Example Question #2413 : Act Math

Example Question #2414 : Act Math

Example Question #2 : How To Divide Rational Expressions

Example Question #3 : How To Divide Rational Expressions

Example Question #2411 : Act Math

Example Question #2 : How To Multiply Rational Expressions

Example Question #2 : How To Multiply Rational Expressions

Example Question #23 : Rational Expressions

All ACT Math Resources

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