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Algebra 1 : Algebra 1

Study concepts, example questions & explanations for Algebra 1

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Example Questions

Example Question #535 : Variables

Simplify:

$\frac{27a^{3}bc}{9a^{5}b^{7}}*\frac{20b^{4}c^{2}}{16a}$

Possible Answers:

$\frac{15}{4a^{3}b^{2}c^{3}}$

$\frac{15a^{\frac{1}{2}}b^{\frac{5}{7}}c^{3}}{4}$

$\frac{15a^{3}b^{2}c^{3}}{4}$

$\frac{4a^{3}b^{2}}{15c^{3}}$

$\frac{15c^{3}}{4a^{3}b^{2}}$

Correct answer:

$\frac{15c^{3}}{4a^{3}b^{2}}$

Explanation:

First, in the first quotient, divide 27 and 9 by the GCF (9). In the second quotient, divide 20 and 16 by the GCF (4).

$\frac{27a^{3}bc}{9a^{5}b^{7}}*\frac{20b^{4}c^{2}}{16a}=\frac{3a^{3}bc}{a^{5}b^{7}}*\frac{5b^{4}c^{2}}{4a}$

Second, organize the variables so like terms are together in the numerator and in the denominator.

$\frac{3a^{3}bc}{a^{5}b^{7}}*\frac{5b^{4}c^{2}}{4a}=\frac{3*5*a^{3}bb^{4}cc^{2}}{4a^{5}ab^{7}}$

Third, multiply the integers and use the rules of exponents to combine the following terms: $a^{5}$ and , and $b^{4}$ , and and $c^{2}$ . Remember the following exponent rule: $x^{a}*x^{b}=x^{a+b}$ ,

$\frac{3*5*a^{3}bb^{4}cc^{2}}{4a^{5}ab^{7}}=\frac{15a^{3}b^{5}c^{3}}{4a^{6}b^{7}}$

Fourth, use the rules of exponents

$\frac{x^{a}}{x^{b}}=x^{a-b}$

and

$x^{-a}=\frac{1}{x^{a}}$

to further simplify the expression by combining the terms $a^{3}$ and $a^{6}$ and $b^{5}$ and $b^{7}$ .

$\frac{15a^{3}b^{5}c^{3}}{4a^{6}b^{7}}=\frac{15a^{-3}b^{-2}c^{2}}{4}=\frac{15c^{3}}{4a^{3}b^{2}}$

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Example Question #4771 : Algebra 1

Simplify:

$\frac{a^{7}b^{6}c^{4}}{8}*\frac{28b^{3}}{12a^{3}b^{2}}$

Possible Answers:

$\frac{7c^{4}}{6a^{10}b^{11}}$

$\frac{7a^{4}b^{7}c^{4}}{6}$

$\frac{7a^{4}b^{7}c^{4}}{24}$

$\frac{24}{7a^{4}b^{7}c^{4}}$

$\frac{7a^{\frac{7}{3}}b^{\frac{9}{2}}c^{4}}{24}$

Correct answer:

$\frac{7a^{4}b^{7}c^{4}}{24}$

Explanation:

Because 28 and 12, the coefficients in the second quotient, share common factors, you can divide them by the GCF (4). Because the cofficients in the first quotient are technically 1 and 8, you cannot further reduce the 8.

$\frac{a^{7}b^{6}c^{4}}{8}*\frac{28b^{3}}{12a^{3}b^{2}}=\frac{a^{7}b^{6}c^{4}}{8}*\frac{7b^{3}}{3a^{3}b^{2}}$

Organize the variables so like terms are together in the numerator and in the denominator. Then, multiply and in the denominator and use rules of exponents to combine $b^{6}$ and $b^{3}$ in the numerator. Remember the following exponent rule: $x^{a}*x^{b}=x^{a+b}$ ,

$\frac{7a^{7}b^{6}b^{3}c^{4}}{8*3a^{3}b^{2}}=\frac{7a^{7}b^{6}b^{3}c^{4}}{24a^{3}b^{2}}=\frac{7a^{7}b^{9}c^{4}}{24a^{3}b^{2}}$

Use the exponent rule

$\frac{x^{a}}{x^{b}}=x^{a-b}$

to further simplify the expression by combining the terms $a^{7}$ and $a^{3}$ and $b^{9}$ and $b^{2}$ .

$\frac{7a^{7}b^{9}c^{4}}{24a^{3}b^{2}}=\frac{7a^{4}b^{7}c^{4}}{24}$

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Example Question #2 : Simplifying Expressions

Multiply, expressing the product in simplest form:

$\small \frac{15x^{5}}{14y^{3}} \cdot \frac{21y^{2} }{25x^{3}}$

Possible Answers:

$\small \small \small \small \frac{10 x^{2}}{9y }$

$\small \small \small \small \frac{9 y}{10x^{2} }$

$\small \small \small \small \small \frac{10 y}{9x^{2} }$

$\small \small \small \frac{9 x^{2}y}{10 }$

$\small \small \frac{9 x^{2}}{10 y}$

Correct answer:

$\small \small \frac{9 x^{2}}{10 y}$

Explanation:

Cross-cancel the coefficients by dividing both 15 and 25 by 5, and both 14 and 21 by 7:

$\small \small \small \small \frac{15x^{5}}{14y^{3}} \cdot \frac{21y^{2} }{25x^{3}} = \small \small \small \small \frac{3x^{5}}{2y^{3}} \cdot \frac{3y^{2} }{5x^{3}} = \frac{3x^{5} \cdot 3y^{2}}{5x^{3} \cdot 2y^{3}} = \frac{3 \cdot 3 \cdot x^{5} y^{2}}{5\cdot 2\cdot x^{3} y^{3}} = \frac{9 x^{5} y^{2}}{10 x^{3} y^{3}}$

Now use the quotient rule on the variables by subtracting exponents:

$\small \frac{9 x^{5} y^{2}}{10 x^{3} y^{3}} = \frac{9 x^{5-3}}{10 y^{3-2}} = \frac{9 x^{2}}{10 y^{1}} = \frac{9 x^{2}}{10 y}$

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Example Question #4772 : Algebra 1

Simplify the following:

$\frac{2xy^2}{3yz^2} \times \frac{15y^2z}{14x^2y}$

Possible Answers:

$\frac{5xy^2}{7z}$

$\frac{5xz}{7y^2}$

$\frac{5y^2}{7xz}$

$\frac{7y^2}{5x^2z}$

Correct answer:

$\frac{5y^2}{7xz}$

Explanation:

In this problem, you have two fractions being multiplied. You can first simplify the coefficients in the numerators and denominators. You can divide and cancel the 2 and 14 each by 2, and the 3 and 15 each by 3:

$\frac{2xy^2}{3yz^2} \times \frac{15y^2z}{14x^2y} = \frac{xy^2}{yz^2} \times \frac{5y^2z}{7x^2y}$

You can multiply the two numerators and two denominators, keeping in mind that when multiplying like variables with exponents, you simplify by adding the exponents together:

$\frac{xy^2}{yz^2} \times \frac{5y^2z}{7x^2y} = \frac{5xy^4z}{7x^2y^2z^2}$

Any variables that are both in the numerator and denominator can be simplified by subtracting the numerator's exponent by the denominator's exponent. If you end up with a negative exponent in the numerator, you can move the variable to the denominator to keep the exponent positive:

$\frac{5xy^4z}{7x^2y^2z^2} = \frac{5y^2}{7xz}$

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Example Question #4773 : Algebra 1

Simplify:

$\frac{9ab^2c}{4c^3} \times \frac{a^2b^2}{3a^3c}$

Possible Answers:

$\frac{3b^4}{4c^3}$

$\frac{3b^4}{4ac^2}$

$\frac{3a^6b^4}{4c^3}$

$\frac{3a^3b^4}{4c^4}$

Correct answer:

$\frac{3b^4}{4c^3}$

Explanation:

In this problem, you have two fractions being multiplied. You can first divide and cancel the coefficients in the numerators and denominators, by dividing 9 and 3 each by 3:

$\frac{9ab^2c}{4c^3} \times \frac{a^2b^2}{3a^3c} = \frac{3ab^2c}{4c^3} \times \frac{a^2b^2}{a^3c}$

Next you can multiply the two numerators, and multiply the two denominators. Remember that when multiplying like variables with exponents, you add the exponents together:

$\frac{3ab^2c}{4c^3} \times \frac{a^2b^2}{a^3c} = \frac{3a^3b^4c}{4a^3c^4}$

If a variable shows up in both the numerator and denominator, you can simplify by subtracting the numerator's exponent by the denominator's exponent. If you end up with a negative exponent in the numerator, you can move the variable and exponent to the denominator to make it positive:

$\frac{3b^4}{4c^3}$

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Example Question #3472 : Algebra Ii

Simplify the following:

$\frac{10x^2y^3z}{7xz^3} \times \frac{21xyz^2}{5y}$

Possible Answers:

6x^2y^3z

$\frac{6xy^2}{z}$

$\frac{6x}{y^2z}$

6x^2y^3

Correct answer:

6x^2y^3

Explanation:

In this problem, you have two fractions being multiplied. You can first divide and cancel the coefficients in the numerators and denominators, by dividing 10 and 5 each by 5 and dividing 21 and 7 each by 7:

$\frac{10x^2y^3z}{7xz^3} \times \frac{21xyz^2}{5y} = \frac{2x^2y^3z}{xz^3} \times \frac{3xyz^2}{y}$

Next you can multiply the two numerators, and multiply the two denominators. Remember that when multiplying like variables with exponents, you add the exponents together:

$\frac{2x^2y^3z}{xz^3} \times \frac{3xyz^2}{y} = \frac{6x^3y^4z^3}{xyz^3}$

If a variable shows up in both the numerator and denominator, you can simplify by subtracting the numerator's exponent by the denominator's exponent:

$\frac{6x^3y^4z^3}{xyz^3} = 6x^2y^3$

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Example Question #4771 : Algebra 1

Simplify the following:

$\frac{2a^3bc}{3a^2bc^4} \times \frac{15ab^3c^2}{5a^3b^5c^2}$

Possible Answers:

$\frac{2}{ab^2c^3}$

$\frac{2c^3}{ab^2}$

$\frac{2ab}{c^3}$

2ab^2c^3

Correct answer:

$\frac{2}{ab^2c^3}$

Explanation:

In this problem, you have two fractions being multiplied. You can first divide and cancel the coefficients in the numerators and denominators. The two coefficients in the denominators multiply up to 15, allowing you to divide and cancel those two coefficients with the 15 in the numerator:

$\frac{2a^3bc}{3a^2bc^4} \times \frac{15ab^3c^2}{5a^3b^5c^2} = \frac{2a^3bc}{a^2bc^4} \times \frac{ab^3c^2}{a^3b^5c^2}$

Next you can multiply the two numerators, and multiply the two denominators. Remember that when multiplying like variables with exponents, you add the exponents together:

$\frac{2a^3bc}{a^2bc^4} \times \frac{ab^3c^2}{a^3b^5c^2} = \frac{2a^4b^4c^3}{a^5b^6c^6}$

$\frac{2a^4b^4c^3}{a^5b^6c^6} = \frac{2}{ab^2c^3}$

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Example Question #1964 : Algebra Ii

Multiply. Give the answer in simplest form.

$\left(\frac{3}{2x^2} \right )\cdot\left(\frac{4x^2y}{9} \right )$

Possible Answers:

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

$\frac{12y}{9}$

$\frac{2y}{3}$

2x^4y

Correct answer:

$\frac{2y}{3}$

Explanation:

Multiplying quotients is similar to multiplying fractions, so we multiply straight across to get $\small \frac{12x^2y}{18x^2}$ . From this point, we can simplify. Since $\small \frac{12}{18}=\frac{2}{3}$ and $\small \frac{x^2}{x^2}=x^{2-2}=x^0=1$ , our final answer becomes $\small \frac{2}{3} \cdot 1 \cdot y = \frac{2y}{3}$ .

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Example Question #4775 : Algebra 1

Simplify:

$\frac{30a^{2}b}{c}*\frac{bc^{2}}{5c}$

Possible Answers:

$\frac{6a^{2}b^{2}}{c^{4}}$

$6a^{2}b^{2}$

$6ab^{2}$

$6a^{2}b^{2}c^{4}$

$\frac{6a^{2}b^{2}c^{2}}{c^{2}}$

Correct answer:

$6a^{2}b^{2}$

Explanation:

Combine like terms in the numerator and the denominator. Use the rules of exponents to combine and in the numerator and and in the denominator. Remember that $x^{a}*x^{b}=x^{a+b}$ Then, divide 30 by 5 (the GCF).

$\frac{30a^{2}b*b*c^{2}}{5c*c}=\frac{30a^{2}b^{2}c^{2}}{5c^{2}}=\frac{6a^{2}b^{2}c^{2}}{c^{2}}$

The GCF rule can also be used to remove $c^{2}$ from the numerator and the deonominator. $c^{2}$ goes into $c^{2}$ once.

$\frac{6a^{2}b^{2}c^{2}}{c^{2}}=6a^{2}b^{2}$

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Example Question #12 : How To Multiply Monomial Quotients

Simplify:

$\frac{c^{4}}{4}*\frac{3a^{2}}{b^{2}}$

Possible Answers:

$\frac{3a^{2}c^{4}}{4b^{2}}$

$\frac{3a^{2}b^{2}c^{4}}{4}$

$\frac{4b^{2}}{3a^{2}c^{4}}$

$\frac{3}{4}a^{2}b^{2}c^{4}$

$\frac{4a^{2}b^{2}c^{4}}{3}$

Correct answer:

$\frac{3a^{2}c^{4}}{4b^{2}}$

Explanation:

Since there are no like terms in the numerator or in the denominator, you can only combine ther terms on the numerator and denominator so that they are in one quotient. You cannot further combine the variables because each variable is represented by a different letter. You cannot further reduce the integers because they do not have a common factor.

$\frac{c^{4}}{4}*\frac{3a^{2}}{b^{2}}=\frac{3a^{2}c^{4}}{4b^{2}}$

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Algebra 1 : Algebra 1

Study concepts, example questions & explanations for Algebra 1

All Algebra 1 Resources

Example Questions

Example Question #535 : Variables

Example Question #4771 : Algebra 1

Example Question #2 : Simplifying Expressions

Example Question #4772 : Algebra 1

Example Question #4773 : Algebra 1

Example Question #3472 : Algebra Ii

Example Question #4771 : Algebra 1

Example Question #1964 : Algebra Ii

Example Question #4775 : Algebra 1

Example Question #12 : How To Multiply Monomial Quotients

All Algebra 1 Resources

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