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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 : Solving Equations

Solve the equation for y

5y+27-5z=30

Possible Answers:

$z=\frac{3-5y}{5}$

5y-5z=3

5y=3+5z

$y=\frac{3+5z}{5}$

y=3+5z

Correct answer:

$y=\frac{3+5z}{5}$

Explanation:

5y+27-5z=30

First subtract 27 from both sides of the equation

5y-5z=3

Add 5z to both sides of the equation

5y=3+5z

Lastly, divide both sides by 5 to get the y by itself

$y=\frac{3+5z}{5}$

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Example Question #1 : Solving Equations

Solve the equation: $\frac{2}{3x}+4=\frac{1}{2x}$

Possible Answers:

$\frac{1}{12}$

$\frac{1}{48}$

$-\frac{1}{12}$

$-\frac{1}{24}$

$\frac{1}{24}$

Correct answer:

$-\frac{1}{24}$

Explanation:

To isolate the x-variable, we can multiply both sides by the least common denominator.

The least common denominator is . This will eliminate the fractions.

$6x(\frac{2}{3x}+4)=\frac{1}{2x} \cdot 6x$

4+24x = 3

Subtract 4 on both sides.

4+24x -4= 3-4

24x=-1

Divide by 24 on both sides.

$\frac{24x}{24}=\frac{-1}{24}$

The answer is: $-\frac{1}{24}$

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Example Question #3 : Solving Equations

Solve the equation: $\frac{1}{3}x-\frac{1}{4}x = \frac{1}{5}$

Possible Answers:

$\frac{7}{60}$

$\frac{12}{5}$

$\frac{10}{3}$

$\frac{24}{5}$

$\frac{13}{60}$

Correct answer:

$\frac{12}{5}$

Explanation:

Find the least common denominator of both sides of the equation, and multiply it on both sides.

The LCD is 60.

$60(\frac{1}{3}x-\frac{1}{4}x) = \frac{1}{5}\times 60$

20x-15x = 12

Combine like-terms on the left.

5x=12

Divide by 5 on both sides.

$\frac{5x}{5}=\frac{12}{5}$

The answer is: $\frac{12}{5}$

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Example Question #1 : Solving Equations

Solve the equation: 8(x-3) = 16-x

Possible Answers:

$\frac{40}{7}$

$\frac{8}{9}$

$\frac{40}{9}$

Correct answer:

$\frac{40}{9}$

Explanation:

Distribute the eight through both terms of the binomial.

8x-24 = 16-x

Add on both sides.

8x-24 +x= 16-x+x

9x-24 = 16

Add 24 on both sides.

9x-24+24 = 16+24

9x=40

Divide by 9 on both sides.

$\frac{9x}{9}=\frac{40}{9}$

The answer is: $\frac{40}{9}$

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Example Question #5 : Solving Equations

Solve the equation: -2+7x = 8x-6

Possible Answers:

$-\frac{4}{15}$

$\frac{4}{15}$

$-\frac{8}{15}$

Correct answer:

Explanation:

Subtract from both sides.

-2+7x-7x = 8x-6-7x

-2=x-6

Add 6 on both sides.

-2+6=x-6+6

x=4

The answer is:

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Example Question #6 : Solving Equations

Solve $\sqrt{x^2 + 8} = 2\sqrt{2x-1}$

Possible Answers:

$x= \sqrt{3}, \sqrt{6}$

x=-3, 6

x=2, 6

x=3, 9

x=7

Correct answer:

x=2, 6

Explanation:

First, we want to get everything inside the square roots, so we distribute the :

$\sqrt{x^2 + 8} = \sqrt{4}\sqrt{2x-1}$

$\sqrt{x^2 + 8} = \sqrt{4(2x-1)}$

$\sqrt{x^2 + 8} = \sqrt{8x-4}$

Now we can clear our the square roots by squaring each side:

$(\sqrt{x^2 + 8})^{2} = (\sqrt{8x-4})^2$

x^2 + 8 = 8x-4

Now we can simplify by moving everything to one side of the equation:

x^2 -8x+ 12 = 0

Factoring will give us:

(x-6)(x-2)= 0

So our answers are:

x=2, 6

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Example Question #2 : Solving Equations

Solve $6-\sqrt{x}=11$

Possible Answers:

No solution.

Correct answer:

No solution.

Explanation:

Begin by gathering all the constants to one side of the equation:

$-\sqrt{x}=5$

Now multiply by :

$\sqrt{x}=-5$

And finally, square each side:

x=25

This might look all fine and dandy, but let's check our solution by plugging it in to the original equation:

$6-\sqrt{25}=11$

6-5=11

$1\neq 11$

So our solution is invalid, and the problem doesn't have a solution.

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Example Question #11 : Single Variable Algebra

Solve the equation: 3x-2 = 50

Possible Answers:

$\frac{52}{3}$

Correct answer:

$\frac{52}{3}$

Explanation:

Add two on both sides.

3x-2+2 = 50+2

3x = 52

Divide by three on both sides.

$\frac{3x }{3}=\frac{ 52}{3}$

The answer is: $\frac{52}{3}$

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Example Question #12 : Single Variable Algebra

Solve: $\frac{2}{3}x = \frac{5}{2}$

Possible Answers:

$\frac{4}{15}$

$\frac{10}{3}$

$\frac{3}{5}$

$\frac{15}{4}$

$\frac{5}{3}$

Correct answer:

$\frac{15}{4}$

Explanation:

To isolate the x-variable, multiply both sides by the coefficient of the x-variable.

$\frac{2}{3}x \cdot \frac{3}{2} = \frac{5}{2}\cdot\frac{3}{2}$

The answer is: $\frac{15}{4}$

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Example Question #172 : Sat Subject Test In Math Ii

Give the solution set of the following rational equation:

$\frac{ x^{2}-4x}{x^{2}-2x} = 2$

Possible Answers:

$\left \{ 2, 4 \right \}$

$\left \{ 0, 2, 4 \right \}$

$\left \{ 0\right \}$

No solution

$\left \{ 0, 4 \right \}$

Correct answer:

No solution

Explanation:

Multiply both sides of the equation by $x^{2}-2x$ to eliminate the fraction:

$\frac{ x^{2}-4x}{x^{2}-2x} = 2$

$\frac{ x^{2}-4x}{x^{2}-2x}\cdot (x^{2}-2x) = 2 \cdot (x^{2}-2x)$

$x^{2}-4x = 2 x^{2}-4x$

Subtract $x^{2}-4x$ from both sides:

$x^{2}-4x- (x^{2}-4x) = 2 x^{2}-4x - (x^{2}-4x)$

$2 x^{2} - x^{2} -4x + 4x = 0$

$x^{2} = 0$

The only possible solution is x = 0 , However, if this is substituted in the original equation, the expression at left is undefined, as seen here:

$\frac{ x^{2}-4x}{x^{2}-2x} = 2$

$\frac{ 0^{2}-4 \cdot 0}{0^{2}-2 \cdot 0} = 2$

$\frac{ 0}{ 0} = 2$

An expression with a denominator of 0 has an undefined value, so this statement is false. The equation has no solution.

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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 : Solving Equations

Example Question #1 : Solving Equations

Example Question #3 : Solving Equations

Example Question #1 : Solving Equations

Example Question #5 : Solving Equations

Example Question #6 : Solving Equations

Example Question #2 : Solving Equations

Example Question #11 : Single Variable Algebra

Example Question #12 : Single Variable Algebra

Example Question #172 : Sat Subject Test In Math Ii

All SAT II Math II Resources

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