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SAT II Math II : Single-Variable Algebra

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Example Questions

Example Question #41 : Single Variable Algebra

Solve the inequality: -3x+9>-15

Possible Answers:

x>2

x<-2

x<2

x<-8

x<8

Correct answer:

x<8

Explanation:

Subtract nine from both sides.

-3x+9-9>-15-9

-3x>-24

Divide by negative 3 on both sides. We will need to switch the sign.

$\frac{-3x}{-3}>\frac{-24}{-3}$

The answer is: x<8

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Example Question #17 : Solving Inequalities

Solve the inequality: $3x-7 \leq -14$

Possible Answers:

$x\leq -7$

$x\leq \frac{7}{3}$

$x\geq -\frac{7}{3}$

$x\leq -\frac{7}{3}$

$x\leq7$

Correct answer:

$x\leq -\frac{7}{3}$

Explanation:

Add seven on both sides.

$3x-7 +7\leq -14+7$

$3x\leq -7$

Divide by three on both sides.

$\frac{3x}{3}\leq \frac{-7}{3}$

The answer is: $x\leq -\frac{7}{3}$

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Example Question #41 : Single Variable Algebra

Solve: $-7 ( x + 9 ) \geq 37 + 3x$

Possible Answers:

$x \geq-7$

$x \leq-10$

$-10 \leq x \leq 10$

$x \geq -10$

$x \leq -10, x\geq10$

Correct answer:

$x \leq-10$

Explanation:

First, we distribute the through the equation:

$-7x - 63 \geq 37 + 3x$

Now, we collect and combine terms:

$10x \leq -100$

$x \leq-10$

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Example Question #41 : Single Variable Algebra

Decrease 8d - 35 by 40%. Which of the following will this be equal to?

Possible Answers:

4.8d-35

4.8 d - 21

3.2d-21

8d-21

3.2d - 14

Correct answer:

4.8 d - 21

Explanation:

A number decreased by 40% is equivalent to 100% of the number minus 40% of the number. This is taking 60% of the number, or, equivalently, multiplying it by 0.6.

Therefore, 8d - 35 decreased by 40% is 0.6 times this, or

$0.6 \left (8d - 35 \right ) = 0.6 \cdot 8d - 0.6 \cdot 35 = 4.8 d - 21$

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Example Question #42 : Single Variable Algebra

Simplify:

$\left ( 3x^{-3}\right )^{-3}$

Possible Answers:

$27x^{9}$

$\frac{1}{27x^{9}}$

$\frac{ x^{9} }{27}$

$-27x^{9}$

$\frac{27}{ x^{9} }$

Correct answer:

$\frac{ x^{9} }{27}$

Explanation:

$\left ( 3x^{-3}\right )^{-3}$

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

$= \left ( \frac{x^{3}}{3}\right )^{3}$

$= \frac{\left (x^{3} \right )^{3}}{3^{3}}$

$= \frac{ x^{3 \cdot 3} }{27}$

$= \frac{ x^{9} }{27}$

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Example Question #42 : Single Variable Algebra

Assume all variables assume positive values.

Simplify:

$\left ( \frac{y^{-3}}{x^{-2}} \right )^{2}$

Possible Answers:

$\frac{1}{x^{4}y^{6}}$

$\frac{x^{4}}{y^{6}}$

$\frac{y^{6}}{x^{4}}$

$\frac{x^{4}}{y^{9}}$

$\frac{y^{9}}{x^{4}}$

Correct answer:

$\frac{x^{4}}{y^{6}}$

Explanation:

$\left ( \frac{y^{-3}}{x^{-2}} \right )^{2}$

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

$= \frac{y^{-6}}{x^{-4}}$

$= y^{-6} \div x^{-4}$

$= \frac{1}{y^{6}} \div \frac{1}{ x^{4}}$

$= \frac{1}{y^{6}} \cdot x^{4}$

$= \frac{x^{4}}{y^{6}}$

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Example Question #1 : Simplifying Expressions

Assume all variables assume positive values. Simplify:

$6x^{0}+ 7y^{0}+ 8z^{0}$

Possible Answers:

The expression is already simplified.

The expression is undefined.

6x + 7y + 8z

21xyz

Correct answer:

Explanation:

Any nonzero expression raised to the power of 0 is equal to 1. Therefore,

$6x^{0}+ 7y^{0}+ 8z^{0}$

$= 6 \cdot 1 + 7 \cdot 1 + 8 \cdot 1$

= 6 + 7 + 8 = 21

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Example Question #192 : Sat Subject Test In Math Ii

Simplify the following expression: xyz(xy-xz)

Possible Answers:

x^2y^2z^2-x^2yz^2

2xyz-2xz

x^2y^2z-x^2yz^2

x^2y^2z-x^2yz

2x^2y^2z

Correct answer:

x^2y^2z-x^2yz^2

Explanation:

Distribute the outer term through both terms in the parentheses.

xyz(xy-xz) = xyz(xy) - xyz(xz)

Multiply each term.

x^2y^2z-x^2yz^2

There are no like-terms.

The answer is: x^2y^2z-x^2yz^2

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Example Question #2 : Simplifying Expressions

Simplify the expression: 6(2x-3)-2(-x-3)

Possible Answers:

14x-24

12x-12

10x-9

14x-12

10x+9

Correct answer:

14x-12

Explanation:

Distribute the integers through the binomials.

12x-18 +2x+6

Combine like-terms.

The answer is: 14x-12

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Example Question #3 : Simplifying Expressions

Simplify x - 5 - (2 - x) .

Possible Answers:

x+3

2x - 7

x^2 - 7

Correct answer:

2x - 7

Explanation:

We can start by distributing the negative sign in the parentheses term:

x - 5 - 2 + x

Now we can combine like terms. The constants go together, and the variables go together:

2x - 7

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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 #41 : Single Variable Algebra

Example Question #17 : Solving Inequalities

Example Question #41 : Single Variable Algebra

Example Question #41 : Single Variable Algebra

Example Question #42 : Single Variable Algebra

Example Question #42 : Single Variable Algebra

Example Question #1 : Simplifying Expressions

Example Question #192 : Sat Subject Test In Math Ii

Example Question #2 : Simplifying Expressions

Example Question #3 : Simplifying Expressions

All SAT II Math II Resources

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