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Algebra II : Solving Absolute Value Equations

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Example Question #31 : Absolute Value

Which of the following is in the set of all possible values for y in the equation $3\left | 5-y\right |+8=29$ ?

Possible Answers:

The empty set.

$\left \{-2, 12 \right \}$

$\left \{ 12\right \}$

$\left \{ 2\right \}$

$\left \{ -12,2\right \}$

Correct answer:

$\left \{-2, 12 \right \}$

Explanation:

First isolate the absolute value:

$3\left | 5-y\right |+8=29$

$3\left | 5-y\right |=21$

$\left | 5-y\right |=7$

There are two possible values for 5-y : 7 and -7.

5-y=7 or 5-y=-7

y=-2 or y=12

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Example Question #2152 : Mathematical Relationships And Basic Graphs

What is the set of all values for x that satisfy the equation $-2\left | 2x+5\right |=22$ ?

Possible Answers:

$\left \{ -8,3\right \}$

$\left \{ -3, 8\right \}$

The empty set.

$\left \{ -\frac{27}{4},\frac{17}{4}\right \}$

$\left \{ -\frac{17}{4}, \frac{27}{4}\right \}$

Correct answer:

The empty set.

Explanation:

First isolate the absolute value.

$-2\left | 2x+5\right |=22$

$\left | 2x+5\right |=-11$

There are no real numbers for which you can take the absolute value and get a negative number. Therefore, there are no values for x that will satisfy the equation.

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Example Question #4821 : Algebra Ii

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

Possible Answers:

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

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

The empty set.

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

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

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 #4822 : Algebra Ii

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 #4823 : Algebra Ii

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 #4824 : Algebra Ii

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 #4825 : Algebra Ii

Solve the following equation:

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

Possible Answers:

x=0, -6

x=6

x=0, 6

x=0

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

v= 8

v=-5, -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 #4827 : Algebra Ii

Solve for x:

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

Possible Answers:

x=8, 12

x=-8

x=-12

x=-8, -12

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 #4828 : Algebra Ii

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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Algebra II : Solving Absolute Value Equations

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #31 : Absolute Value

Example Question #2152 : Mathematical Relationships And Basic Graphs

Example Question #4821 : Algebra Ii

Example Question #4822 : Algebra Ii

Example Question #4823 : Algebra Ii

Example Question #4824 : Algebra Ii

Example Question #4825 : Algebra Ii

Example Question #2161 : Mathematical Relationships And Basic Graphs

Example Question #4827 : Algebra Ii

Example Question #4828 : Algebra Ii

All Algebra II Resources

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