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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #15 : Solving Equations

Solve for :

$\frac{w+2}{4}+\frac{w}{3}=6$

Possible Answers:

$w=\frac{3}{14}$

$w=\frac{27}{16}$

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

$w=\frac{52}{31}$

$w=\frac{66}{7}$

Correct answer:

$w=\frac{66}{7}$

Explanation:

Step 1: The least common denominator of 4 and 3 is 12. Multiply each term by this LCD:

$12\left ( \frac{w+2}{4} \right )+12\left ( \frac{w}{3}\right )=6(12)$

Step 2: Simplify the fractions:

3w+6+4w=72

Step 3: Isolate :

7w+6=72

7w=66

${\color{Red} w=\frac{66}{7}}$

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Example Question #21 : Solving Equations

A large water tank has a water pipe that can be used to fill the tank in forty-five minutes. It has a drain that can empty the tank in one hour and twenty minutes.

One day, someone left the drain open when filling the tank. The tank was completely full by the time someone realized the error. Which of the following comes closest to the amount of time it took to fill the tank?

Possible Answers:

$2\textup{ hr }30\textup{ min}$

$1\textup{ hr }30\textup{ min}$

$2\textup{ hr }15\textup{ min}$

$1\textup{ hr }45\textup{ min}$

$2\textup{ hr }$

Correct answer:

$1\textup{ hr }45\textup{ min}$

Explanation:

Work problems can be solved by looking at them as rate problems. Therefore, we can look at this problem in terms of tanks per minute, rather than minutes per tank. Let be the number of minutes it took to fill the tank.

The pipe filled the tank at a rate of 45 minutes per tank, or $\frac{1}{45}$ tank per minute; over a period of minutes, it filled $\frac{1}{45} X$ tank.

The drain emptied the tank at a rate of 80 minutes per tank, so we can see this as a drain of $\frac{1}{80}$ tank per minute. We can look at draining as "filling negative tanks" - $-\frac{1}{80}$ tank per minute; over a period of minutes, it "filled" $-\frac{1}{80} X$ tank.

Since their work adds up to one tank filled, We can set up, and solve for in, the equation:

$\frac{1}{45} X -\frac{1}{80} X = 1$

$\left (\frac{1}{45} -\frac{1}{80} \right )X = 1$

Using decimal approximations:

$\left (0.02222 - 0.01250 \right )X = 1$

0.00972 X = 1

$X = 1 \div 0.00972 \approx 103$ minutes, or 1 hour 43 minutes.

Of the given choices, 1 hour 45 minutes is closest.

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Example Question #2341 : Algebra Ii

Solve for x:

$\frac{x}{6} + 4 = 10$

Possible Answers:

Correct answer:

Explanation:

In order to solve for x, first subtract 4 from both sides of the equation:

$\frac{x}{6} = 6$

Then, multiply both sides of the equation by 6:

x = 36

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Example Question #2342 : Algebra Ii

Solve for :

$(\frac{1}{2})x +7=15$

Possible Answers:

x=-4

x=16

x=8

x=4

x=-8

Correct answer:

x=16

Explanation:

$(\frac{1}{2})x=15-7=8$

$x=(\frac{2}{1})\times8=16$

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Example Question #23 : Solving Equations

Solve this system of equations:

$\newline 2x+2y=4 \newline -x-y=-8 \mathor$

Possible Answers:

$\newline x=5 \newline y=4$

$\newline x=3 \newline y=5$

$\newline x=4 \newline y=2$

$\newline x=2 \newline y=-4$

$\newline x=-8 \newline y=-5$

Correct answer:

$\newline x=3 \newline y=5$

Explanation:

Solve both equations for y:

$\newline y = \frac{4+2x}{2} \newline \newline y=-x+8$

Set them equal to each other and solve for :

$\newline -x+8=\frac{4+2x}{2} \newline \newline -x+8=2+x \newline \rightarrow x=3$

Plug x back into either -equation to get

y=5 .

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Example Question #23 : Solving Equations

Solve for :

25=5x-5

Possible Answers:

x=6

x=12

x=5

x=8

Correct answer:

x=6

Explanation:

To solve for x we want to isolate x on one side of the equation and all other numbers on the other side. To do this we start with adding 5 to both sides.

25=5x-5

30=5x

Now we divide by 5 to solve for x.

$\frac{30}{5}=\frac{5x}{5}$

6=x

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Example Question #503 : Basic Single Variable Algebra

Solve for :

5x - 9 = 19 + x

Possible Answers:

Correct answer:

Explanation:

In order to solve for , first add 9 to both sides of the equation:

5x = 28 + x

Then, subtract from both sides of the equation:

4x = 28

Finally, divide both sides of the equation by 4:

x = 7

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Example Question #2343 : Algebra Ii

Solve for :

7x - 6 = 15

Possible Answers:

Correct answer:

Explanation:

Add 6 to both sides of the equation:

7x = 21

Then divide both sides of the equation by 7:

x = 3

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Example Question #24 : Solving Equations

Solve for .

3x-2(x-1) = 4x+3

Possible Answers:

$x = \frac{1}{3}$

$x = -\frac{1}{3}$

x = 3

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

x = -3

Correct answer:

$x = -\frac{1}{3}$

Explanation:

This problem requires simplification, knowledge of order of operations, and ability to isolate variables and solve for .

Our first step is to simplify the left side of the equation. Use the distributive property to simplify

3x-2(x-1) Distribute the (Don't forget the negative!!)

3x-2x+2 Combine like terms

x+2

Now we have....

x+2 = 4x+3

Combine like terms, simplify, and solve for

x+2 = 4x+3 Subtract from both sides

2 = 3x + 3 Subtract from both sides

-1 = 3x Divide by on both sides

-1/3 = x

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Example Question #21 : Solving Equations

Solve for :

4x=-20=-9x+19

Possible Answers:

Correct answer:

Explanation:

In order to solve this equation, we need to move all constants to one side and everything that has an to the other side.

$4x-\underset{+20}{20}=-9x+\underset{+20}{19}$

which becomes

4x=-9x+39

Moving the to the left side results in

$\underset{+9x}{4x}=-\underset{+9x}{9x}+39$

13x = 39

Dividing each side by 13 gives us

$\frac{13x}{13}=\frac{39}{13}=3=x$

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #15 : Solving Equations

Example Question #21 : Solving Equations

Example Question #2341 : Algebra Ii

Example Question #2342 : Algebra Ii

Example Question #23 : Solving Equations

Example Question #23 : Solving Equations

Example Question #503 : Basic Single Variable Algebra

Example Question #2343 : Algebra Ii

Example Question #24 : Solving Equations

Example Question #21 : Solving Equations

All Algebra II Resources

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