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Algebra II : Basic Single-Variable Algebra

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Example Questions

Example Question #1992 : Algebra Ii

Simplify the expression: 2a(a-3)+a^2-3a

Possible Answers:

a^2-6a

3a^2-9a

a^2+6a

3a^2+9a

3a^2-27a

Correct answer:

3a^2-9a

Explanation:

Simplify the first expression.

2a(a-3) = 2a^2-6a

Rewrite the expression and combine like-terms.

2a(a-3)+a^2-3a= 2a^2-6a+a^2-3a

The answer is: 3a^2-9a

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Example Question #1993 : Algebra Ii

Simplify the expression: 6ab(a^2-b)-ab

Possible Answers:

6ab^3-6ab^2-ab

12a^3b-ab

6a^3b-7ab^2

-13a^3b^3

6a^3b-6ab^2-ab

Correct answer:

6a^3b-6ab^2-ab

Explanation:

Distribute the first term with the binomial.

6ab(a^2-b) = 6ab(a^2)-6ab(b)= 6a^3b-6ab^2

Subtract the last term with this expression. Do not combine the terms as one unit. There are no like-terms, and the terms cannot simplified any further.

The answer is: 6a^3b-6ab^2-ab

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Example Question #151 : Basic Single Variable Algebra

Simplify the expression: $6xy(-2x-2y+\frac{x}{6})$

Possible Answers:

-11x^2y+12xy^2

-11x^2y-12xy^2

-13x^2y-12xy^2

-12x^2y-12xy^2+y

-12x^2y-12xy^2

Correct answer:

-11x^2y-12xy^2

Explanation:

Multiply the first term with every term inside the parentheses.

6xy(-2x) = -12x^2y

6xy(-2y) = -12xy^2

$6xy(\frac{x}{6}) =x^2y$

Sum the terms.

-12x^2y-12xy^2+x^2y

Combine like-terms.

The answer is: -11x^2y-12xy^2

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Example Question #61 : Simplifying Expressions

Simplify the following expression:

$512^\frac{5}{9}-(\frac{1}{16})^\frac{-3}{4}$

Possible Answers:

Correct answer:

Explanation:

First, focus on the first term $512^\frac{5}{9}$ . By the rules for multiplying exponents,

$512^\frac{5}{9}=(512^\frac{1}{9})^5$ .

Since 2^9=512 , $512^\frac{1}{9}=2$ . Hence,

$(512^\frac{1}{9})^5=2^5=32$ .

Now focus on the second term, $(\frac{1}{16})^\frac{-3}{4}$ . Since the exponent is negative, we must rewrite this expression as the reciprocal of the base to the positive $\frac{3}{4}$ power:

$(\frac{1}{16})^\frac{-3}{4}=16^\frac{3}{4}$

By the rules for multiplying exponents,

$16^\frac{3}{4}=(16^\frac{1}{4})^3$

Since 2^4=16 , $16^\frac{1}{4}=2$ . Hence,

$(16^\frac{1}{4})^3=2^3=8$ .

Substituting these simplified terms into the original expression yields

$512^\frac{5}{9}-(\frac{1}{16})^\frac{-3}{4}=32-8=24$ .

Hence, is the correct answer.

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Example Question #1994 : Algebra Ii

Simplify: -4(2x+3)+2(8x-1)

Possible Answers:

14x-8

8x-2

8x-14

8x+14

4x-14

Correct answer:

8x-14

Explanation:

First, use the distributive property:

-8x-12+16x-2

Now simplify by combining your like terms:

-8x+16x=8x

-12-2=-14

Therefore, your final answer is:

8x-14

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Example Question #61 : Simplifying Expressions

Simplify: $7ab^2(a^2-b^{-2})$

Possible Answers:

7a^3b^2-7a

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

a^3b^2-a

7a^3b^2

$7a^3b^2-\frac{7a}{b^4}$

Correct answer:

7a^3b^2-7a

Explanation:

Distribute the outer term with both terms inside the parentheses.

$7ab^2(a^2-b^{-2}) = (7ab^2)(a^2)- (7ab^2)(b^{-2})$

When similar bases are multiplied, we can add the exponents of that base.

Simplify the terms.

$(7ab^2)(a^2)- (7ab^2)(b^{-2}) = 7a^3b^2- 7ab^0$

Any term besides zero raised to the power of zero equals one.

The answer is: 7a^3b^2-7a

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Example Question #151 : Basic Single Variable Algebra

Given the expression $\frac{w}{x}-\frac{w}{y}-\frac{w}{z}$ , what must be the numerator of the simplified form?

Possible Answers:

wyz-wxz-wxy

xyz-wyz-wxz

yz+xz+xy

yz-xz-xy

Correct answer:

wyz-wxz-wxy

Explanation:

To determine the number, we will need to find the least common denominator by multiplying all three denominators together.

(x)(y)(z)=xyz

Simplify the fractions.

$\frac{w}{x}-\frac{w}{y}-\frac{w}{z} = \frac{wyz}{xyz}-\frac{wxz}{xyz}-\frac{wxy}{xyz}$

The answer is: wyz-wxz-wxy

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Example Question #1995 : Algebra Ii

If a=-2 and b=-3 , what does (a+b)^2-(a-b)^2 equal?

Possible Answers:

Correct answer:

Explanation:

Substitute the terms into the expression.

(a+b)^2-(a-b)^2 = ((-2)+(-3))^2-((-2)-(-3))^2

Simplify the terms by order of operation. Start with the inner parentheses.

(-5)^2-(1)^2 = 25-1 = 24

The answer is:

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Example Question #152 : Basic Single Variable Algebra

Simplify: abc^2(ac-bc^2)

Possible Answers:

2abc^2-4bc

2a^2bc^3-2ab^2c^4

2abc^2-bc

a^2bc^3-ab^2c^4

a^2bc^3-2ab^2c^2

Correct answer:

a^2bc^3-ab^2c^4

Explanation:

In order to simplified the expression, we will need to distribute the outer term through the terms inside the parentheses.

abc^2(ac-bc^2)=abc^2(ac)-abc^2(bc^2)

Distribute the terms.

The answer is: a^2bc^3-ab^2c^4

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Example Question #153 : Basic Single Variable Algebra

Simplify: $\frac{\frac{y}{x(x+1)}}{1+\frac{y}{x(x+1)}}$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Simplify the complex fraction by parts.

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

Replace the terms.

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

Simplify the denominator. Convert the integer using the least common denominator, which is x^2+x .

$1+\frac{y}{x^2+x}=\frac{x^2+x}{x^2+x}+\frac{y}{x^2+x} = \frac{x^2+x+y}{x^2+x}$

Replace this term in the denominator.

$\frac{\frac{y}{x^2+x}}{1+\frac{y}{x^2+x}}=\frac{\frac{y}{x^2+x}}{ \frac{x^2+x+y}{x^2+x}}$

Rewrite this using a division sign.

$\frac{\frac{y}{x^2+x}}{ \frac{x^2+x+y}{x^2+x}} = \frac{y}{x^2+x} \div\frac{x^2+x+y}{x^2+x}$

Change the sign to multiplication and take the reciprocal of the second term.

$\frac{y}{x^2+x} \cdot \frac{x^2+x}{x^2+x+y}$

Cancel the common terms.

The answer is: $\frac{y}{x^2+x+y}$

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Algebra II : Basic Single-Variable Algebra

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #1992 : Algebra Ii

Example Question #1993 : Algebra Ii

Example Question #151 : Basic Single Variable Algebra

Example Question #61 : Simplifying Expressions

Example Question #1994 : Algebra Ii

Example Question #61 : Simplifying Expressions

Example Question #151 : Basic Single Variable Algebra

Example Question #1995 : Algebra Ii

Example Question #152 : Basic Single Variable Algebra

Example Question #153 : Basic Single Variable Algebra

All Algebra II Resources

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