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SAT II Math II : Solving Equations

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Example Question #7 : Finding Roots

Solve for :

x^2+2 = 6x-6

Possible Answers:

x=1, 2

x=2,4

x=-2,-4

x=1,-2

Correct answer:

x=2,4

Explanation:

To solve for , you need to isolate it to one side of the equation. You can subtract the from the right to the left. Then you can add the 6 from the right to the left:

x^2+2 = 6x-6

x^2-6x+2 = -6

x^2-6x+8 = 0

Next, you can factor out this quadratic equation to solve for . You need to determine which factors of 8 add up to negative 6:

$(x \ \ \ \ \ \)(x \ \ \ \ \ \) = 0$

(x-2)(x-4)=0

Finally, you set each binomial equal to 0 and solve for :

$x=2 \ \ \ \ \ \ x=4$

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Example Question #1 : Solving Equations

Give the solution set of the equation $x ^{2} - 4x + 11 = 0$ .

Possible Answers:

$x =- \sqrt{7} \pm 2i$

$x = 2 \pm \sqrt{7}$

$x = 2 \pm i \sqrt{7}$

$x = \sqrt{7} \pm 2i$

$x = -2 \pm i \sqrt{7}$

Correct answer:

$x = 2 \pm i \sqrt{7}$

Explanation:

Use the quadratic formula with a = 1, b = -4, c = 11 :

$x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

$x = \frac{-(-4) \pm \sqrt{(-4)^{2}-4 (1)(11)}}{2 (1)}$

$x = \frac{4 \pm \sqrt{16-44}}{2 }$

$x = \frac{4 \pm \sqrt{-28}}{2 }$

$x = \frac{4 \pm i \sqrt{28}}{2 }$

$x = \frac{4 \pm i \cdot \sqrt{4} \cdot \sqrt{7}}{2 }$

$x = \frac{4 \pm 2i \sqrt{7}}{2 }$

$x = 2 \pm i \sqrt{7}$

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Example Question #2 : Solving Equations

A large tub has two faucets. The hot water faucet, if turned all the way up, can fill the tub in 15 minutes; the cold water faucet can do the same in 9 minutes. Which of the following responses is closest to the time it takes to fill the tub if both faucets are turned all the way up?

Possible Answers:

$6\textup{ min}$

$12 \textup{ min}$

$4\textup{ min}$

$8\textup{ min}$

$10\textup{ min}$

Correct answer:

$6\textup{ min}$

Explanation:

Work problems can be solved by looking at them as rate problems.

The hot faucet can fill up the tub at a rate of 15 minutes per tub, or $\frac{1}{15}$ tub per minute. The cold faucet, similarly, can fill up the tub at a rate of 9 minutes per tub, or $\frac{1}{9}$ tub per minute.

Suppose the tub fills up in minutes. Then, at the end of this time, the hot faucet has filled up $\frac{1}{15} X$ tub, and the cold faucet has filled up $\frac{1}{9} X$ tub, for a total of one tub. We can set up this equation and solve for :

$\frac{1}{15} X + \frac{1}{9} X = 1$

$45 \cdot \left (\frac{1}{15} X + \frac{1}{9} X \right )= 45 \cdot 1$

$45 \cdot \frac{1}{15} X +45 \cdot \frac{1}{9} X = 45$

3 X +5 X = 45

8 X = 45

$8 X \div 8 = 45 \div 8$

X = 5.625

Of the five choices, 6 minutes comes closest.

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Example Question #11 : Single Variable Algebra

Solve the equation for y

5y+27-5z=30

Possible Answers:

y=3+5z

5y-5z=3

5y=3+5z

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

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

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 #4 : Solving Equations

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

Possible Answers:

$-\frac{1}{12}$

$\frac{1}{24}$

$\frac{1}{48}$

$-\frac{1}{24}$

$\frac{1}{12}$

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 #161 : Sat Subject Test In Math Ii

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

Possible Answers:

$\frac{10}{3}$

$\frac{24}{5}$

$\frac{12}{5}$

$\frac{13}{60}$

$\frac{7}{60}$

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:

$\frac{8}{9}$

$\frac{40}{9}$

$\frac{40}{7}$

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 #11 : Single Variable Algebra

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

Possible Answers:

$\frac{4}{15}$

$-\frac{8}{15}$

$-\frac{4}{15}$

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 #1 : Solving Equations

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

Possible Answers:

x=3, 9

x=-3, 6

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

x=2, 6

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 #1 : 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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SAT II Math II : Solving Equations

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

All SAT II Math II Resources

Example Questions

Example Question #7 : Finding Roots

Example Question #1 : Solving Equations

Example Question #2 : Solving Equations

Example Question #11 : Single Variable Algebra

Example Question #4 : Solving Equations

Example Question #161 : Sat Subject Test In Math Ii

Example Question #1 : Solving Equations

Example Question #11 : Single Variable Algebra

Example Question #1 : Solving Equations

Example Question #1 : Solving Equations

All SAT II Math II Resources

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