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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

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10 Diagnostic Tests 630 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #33 : Absolute Value

Which of the following is the set of all values for that satisfy the equation $5\left |2-3x \right |-33=-8$ ?

Possible Answers:

$\left \{ -5,5\right \}$

The empty set.

$\left \{ -1,\frac{7}{3}\right \}$

$\left \{ -\frac{7}{3}, 1\right \}$

$\left \{ -\frac{31}{3}\right \}$

Correct answer:

$\left \{ -1,\frac{7}{3}\right \}$

Explanation:

First isolate the absolute value.

$5\left |2-3x \right |-33=-8$

$5\left |2-3x \right |=25$

$\left |2-3x \right |=5$

The two possible values for 2-3x are 5 and -5.

2-3x=5 or 2-3x=-5

x=-1 or $x=\frac{7}{3}$

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Example Question #31 : Absolute Value

Solve for .

$2\left | x\right |+14=16$

Possible Answers:

$\pm1$

Correct answer:

$\pm1$

Explanation:

When dealing with absolute value equations, we need to deal with negative values as well.

$2\left | x\right |+14=16$ Subtract on both sides.

$2\left | x\right |=2$ Divide on both sides.

$\left | x\right |=1$

$x=\pm1$

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Example Question #35 : Absolute Value

Solve for .

$\left | x+4\right |=16$

Possible Answers:

12, -20

$\pm12$

$\pm20$

Correct answer:

12, -20

Explanation:

When dealing with absolute value equations, we need to deal with negative values as well.

x+4=16 Subtract on both sides.

x=12

-(x+4)=16 Distribute the negative sign to each term in the parenthesEs.

-x-4=16 Add on both sides.

-x=20 Divide both sides by .

x=-20

Answers are 12, -20

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Example Question #36 : Absolute Value

Solve for .

$\left | x+18\right |-15=29$

Possible Answers:

26,-62

$\pm26$

$\pm62$

Correct answer:

26,-62

Explanation:

When dealing with absolute value equations, we need to deal with negative values as well.

$\left | x+18\right |-15=29$ Add on both sides.

$\left | x+18\right |=44$

x+18=44 Subtract on both sides.

x=26

-(x+18)=44 Distribute the negative sign to each term in the parentheses.

-x-18=44 Add on both sides.

-x=62 Divide on both sides.

x=-62

Answers are 26, -62

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Example Question #37 : Absolute Value

Solve the following equation:

$\left | x-3 \right |-5=-2$

Possible Answers:

x=6

x=0, -6

x=0

x=0, 6

Correct answer:

x=0, 6

Explanation:

To solve absolute value equations, first we must isolate the absolute value:

$\left | x-3 \right |=-2+5=3$

Now that the absolute value is the only thing on the left hand side of the equation, we can solve for x.

Keep in mind that the absolute value makes whatever is inside of it a positive value. This means that x-3 and -(x-3) are both valid in solving the equation.

Setting both of these equal to 3 and solving for x, we get

x-3=3

-x+3=3

x=0, 6

As a check, plug each of these solutions back into the original equation and see if both solutions are valid!

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Example Question #2161 : Mathematical Relationships And Basic Graphs

$\left | 2v-3 \right |=13$

Possible Answers:

v=-5, -8

v=-5, 8

v=-5

v= 8

Correct answer:

v=-5, 8

Explanation:

To solve an absolute value equation, one must first rewrite the problem two different ways.

One is

2v-3=13

and the other is

2v-3=-13 .

Then, solve each equation for v.

Your answers are 8 and -5.

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Example Question #39 : Absolute Value

Solve for x:

$\left | x+10 |-10=-8$

Possible Answers:

x=-12

x=-8, -12

x=8, 12

x=-8

Correct answer:

x=-8, -12

Explanation:

To solve for x, we first must isolate the absolute value on one side of the equation:

$\left | x+10\right |=2$

Remember that an absolute value makes whatever is inside of it positive, so this equation is valid for both what is given inside the absolute value, and also the negative of what is inside:

x+10=2

-(x+10)=2

x=-8

-x-10=2

x=-12

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Example Question #33 : Solving Absolute Value Equations

Solve for x:

$\left | x-2\right |+10x=20$

Possible Answers:

x=0

x=2

x=-2

x=1

Correct answer:

x=2

Explanation:

To solve for x, we first must isolate the absolute value:

$\left | x-2\right |=20-10x$

The absolute value makes whatever is inside of it positive, so the equation is valid for what is given, as well as for the negative of what is inside:

x-2=20-10x

-x+2=20-10x

Now, solve for x for both equations:

9x=18

x=2

Both equations get us this result.

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Example Question #4829 : Algebra Ii

Solve: $\left | 3x-4\right |=49$

Possible Answers:

$-15,\frac{53}{3}$

$-\frac{53}{3},15$

$15,\frac{53}{3}$

$-\frac{53}{3},-15$

$\textup{The answer is not given.}$

Correct answer:

$-15,\frac{53}{3}$

Explanation:

Break up the absolute value into its positive and negative components.

The equations are:

3x-4=49

-(3x-4)=49

Solve the first equation. Add 4 on both sides.

3x-4=49

3x-4+4=49+4

3x=53

Divide by three on both sides.

$\frac{3x}{3}=\frac{53}{3}$

The first solution is: $\frac{53}{3}$

Solve the second equation. Divide by negative one on both sides to move the negative from left to right.

$\frac{-(3x-4)}{-1}=\frac{49}{-1}$

3x-4=-49

Add 4 on both sides.

3x-4+4=-49+4

3x=-45

Divide by three on both sides.

$\frac{3x}{3}=\frac{-45}{3}$

x=-15

The solutions are: $-15,\frac{53}{3}$

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Example Question #41 : Solving Absolute Value Equations

Solve:

$\left | x+5\right |-4=8$

Possible Answers:

x=-17, 7

x=7

x=-17

x=0

Correct answer:

x=-17, 7

Explanation:

To solve, we must first isolate the absolute value on one side:

$\left | x+5\right |=12$

Next, we must remember that the equation holds true for both the positive and negative of what is inside the absolute value sign, because the absolute value makes whatever is inside of it positive:

x+5=12

-(x+5)=12

Solving for x, we get

x=12-5=7

x+5=-12

x=-17

x=-17, 7

One can plug the solutions back into the original equation to confirm.

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #33 : Absolute Value

Example Question #31 : Absolute Value

Example Question #35 : Absolute Value

Example Question #36 : Absolute Value

Example Question #37 : Absolute Value

Example Question #2161 : Mathematical Relationships And Basic Graphs

Example Question #39 : Absolute Value

Example Question #33 : Solving Absolute Value Equations

Example Question #4829 : Algebra Ii

Example Question #41 : Solving Absolute Value Equations

All Algebra II Resources

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