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SAT II Math II : Single-Variable Algebra

Study concepts, example questions & explanations for SAT II Math II

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All SAT II Math II Resources

2 Diagnostic Tests 130 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Simplifying Expressions

Simplify -(x + y) + 4x + 2y .

Possible Answers:

3(x+y)

3xy

3x + y

3x-y

5x+3y

Correct answer:

3x + y

Explanation:

First, we can distribute the negative sign through the parentheses term:

-x - y + 4x + 2y

Now we gather like terms. Remember, you can't gather different variables together. The 's and 's will still be separate terms:

3x + y

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Example Question #4 : Simplifying Expressions

Simplify -(x + 2y + 3) - 3x + y .

Possible Answers:

2x +3y +3

4x +y +3

-3x^2 -2y^2 - 3

2x +3y -3

-4x - y - 3

Correct answer:

-4x - y - 3

Explanation:

Start by distributing the negative sign through the parentheses term:

-x - 2y - 3 - 3x + y

Now combine like terms. Each variable can't be combined with different variables:

-4x - y - 3

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Example Question #5 : Simplifying Expressions

Simplify $(\sqrt{x^2})^2$

Possible Answers:

$x^{\frac{1}{4}}$

$\sqrt{x^4}$

x^2

$\sqrt{x^3}$

Correct answer:

x^2

Explanation:

A square root is the inverse of squaring a term, so they cancel each other out:

$(\sqrt{x^2})^2=x^2$

From there, there's nothing left to simplify.

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Example Question #53 : Single Variable Algebra

Simplify $\sqrt{x^{11}}$ .

Possible Answers:

$(\sqrt{x})^5$

$\sqrt[5]{x}$

$5\sqrt{x}$

$x^{11}$

$x^5 \cdot \sqrt{x}$

Correct answer:

$x^5 \cdot \sqrt{x}$

Explanation:

To begin, let's rewrite the equation so the square root is a fraction in the exponent:

$x^{\frac{11}{2}}$

From here, we can simplify the exponent:

$x^5 \cdot x^{\frac{1}{2}}$

Now we change the exponent fraction back into a square root:

$x^5 \cdot \sqrt{x}$

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Example Question #54 : Single Variable Algebra

Simplify $(\sqrt{4x^2})\cdot (\sqrt{25x})^2$ .

Possible Answers:

100x^2

50x^2

25x +2x

$5 \sqrt{2x}$

50x^3

Correct answer:

50x^2

Explanation:

For the first square root, each term inside has a natural solution. We can take the square root of each term individually because they are multiplied, and then combine them again:

$2x \cdot (\sqrt{25x})^2$

For the second square root, we remember that the square root and a square cancel each other out, and we're left with just the term inside:

$2x \cdot 25x$

We finish by multiplying the terms together:

50x^2

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Example Question #55 : Single Variable Algebra

Simplify 2x^2 -5x(1 - 8x) .

Possible Answers:

2x^2-13x

42x^2 -5x

2x^2-13x-1

-38x^2 -5x

-11x^2

Correct answer:

42x^2 -5x

Explanation:

We start by distributing the -5x term through the parentheses:

2x^2 -5x + 40x^2

Now we combine like terms. Remember, we can't add variables if they have different exponent terms:

42x^2 -5x

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Example Question #14 : Simplifying Expressions

Simplify 2x^2+7x(-8x- 10) .

Possible Answers:

-54x^2- 70x

2x^2 -14x

56x^2 - 70x

-124x^3

2x^2 -x-10

Correct answer:

-54x^2- 70x

Explanation:

Start by distributing the term:

2x^2-56x^2- 70x

Now combine like terms. Remember, if a variable has a different exponent, you can't add them:

-54x^2- 70x

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Example Question #15 : Simplifying Expressions

Simplify -10x(9x+ 4) - 5x^2 .

Possible Answers:

85x^2 +40x

-135x^3

95x^2+40x

-95x^2- 40x

-90x^2 -45x

Correct answer:

-95x^2- 40x

Explanation:

Start by distributing the -10x term:

-90x^2- 40x - 5x^2

Now collect like terms. Remember, you can't add or subtract variables that have different exponents:

-95x^2- 40x

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Example Question #53 : Single Variable Algebra

Simplify -x^2 -6x(8 + 6x) .

Possible Answers:

-x^2 -36x -48

- 37x^2-48x

(-x-8)(x+6)

35x^2+48x

-75x^3

Correct answer:

- 37x^2-48x

Explanation:

Start by distributing the -6x term:

-x^2 -48x - 36x^2

Now combine like terms. Remember, you can't add or subtract variables with different exponents:

- 37x^2-48x

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Example Question #56 : Single Variable Algebra

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

Possible Answers:

2x^2

$-\frac{1}{6}x^2$

$-\frac{7}{6}x^2$

-2x^2

$\frac{1}{6}x^2$

Correct answer:

$-\frac{1}{6}x^2$

Explanation:

Multiply the right terms.

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

Convert to common denominators.

$-\frac{2}{3}x^2 + \frac{1}{2}x^2 = -\frac{4}{6}x^2 + \frac{3}{6}x^2 = -\frac{1}{6}x^2$

The answer is: $-\frac{1}{6}x^2$

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All SAT II Math II Resources

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SAT II Math II : Single-Variable Algebra

Study concepts, example questions & explanations for SAT II Math II

All SAT II Math II Resources

Example Questions

Example Question #1 : Simplifying Expressions

Example Question #4 : Simplifying Expressions

Example Question #5 : Simplifying Expressions

Example Question #53 : Single Variable Algebra

Example Question #54 : Single Variable Algebra

Example Question #55 : Single Variable Algebra

Example Question #14 : Simplifying Expressions

Example Question #15 : Simplifying Expressions

Example Question #53 : Single Variable Algebra

Example Question #56 : Single Variable Algebra

All SAT II Math II Resources

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