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Pre-Algebra : Absolute Value

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Example Questions

Example Question #323 : Operations And Properties

Solve for .

$\left | x+1\right |=3$

Possible Answers:

-4, 2

2, 3

Correct answer:

-4, 2

Explanation:

When taking absolute values, we need to consider both positive and negative values. So, we have two equations. -(x+1)=3, x+1=3

For the left equation, we can switch the minus sign to the other side to get x+1=-3 . When we subtract on both sides, we get x=-4 .

For the right equation, just subtract on both sides, we get x=2 .

x=2, x=-4

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Example Question #324 : Operations And Properties

Solve for .

$\left | 4x\right |=12$

Possible Answers:

$\pm3$

Correct answer:

$\pm3$

Explanation:

When taking absolute values, we need to consider both positive and negative values. So, we have two equations. 4x=12, -(4x)=12

For the left equation, when we divide both sides by , x=3 .

For the right equation, we distribute the negative sign to get -4x=12 . When we divide both sides by , x=-3 .

x=3, x=-3

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Example Question #325 : Operations And Properties

Solve for .

$\left | 2x\right |+5=9$

Possible Answers:

-2, -4

$\pm2$

Correct answer:

$\pm2$

Explanation:

When taking absolute values, we need to consider both positive and negative values. Let's first subtract on both sides. So, we have two equations. 2x=4, -(2x)=4

For the left equation, when we divide both sides by , x=2 .

For the right equation, we distribute the negative sign to get -2x=4 . When we divide both sides by , x=-2 .

x=2, x=-2

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Example Question #14 : Absolute Value

Solve for

$16+4x=4+2\left | x\right |$

Possible Answers:

2,6

-2, -6

Correct answer:

Explanation:

When taking absolute values, we need to consider both positive and negative values. So, we have two equations. 16+4x=4+2x, 16+4x=4-2x

For the left equation, we subtract on both sides and subtract on both sides. We now have 12=-2x . When we divide both sides by , x=-6 .

For the right equation, we subtract on both sides and subtract on both sides. We now have 12=-6x . When we divide both sides by , x=-2 .

Let's double check. When we plug in , both sides aren't equal.

$16+4(-6)=4+2\left | -6\right |$

16-24=4+12

$-8\neq16$

But if we plug in -2, we get both sides equal.

$16+4(-2)=4+2\left | -2\right |$

16-8=4+4

8=8

So is the only answer.

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Example Question #11 : Absolute Value

Solve for

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

Possible Answers:

$\pm2$

No possible answer

Correct answer:

No possible answer

Explanation:

Let's isolate the variable by subtracting both sides by . We have:

$-\left | x\right |=2$ This will be a contradicting expression. Absolute values always generate positive values and since there's a negatie sign in front of it, it will never match a positive value. Therefore no possible answer exist.

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Example Question #332 : Operations And Properties

Solve for .

$\left | x\right |-6=0$

Possible Answers:

$\pm6$

Correct answer:

$\pm6$

Explanation:

When taking absolute values, we need to consider both positive and negative values. So, we have two equations. -x-6=0, x-6=0

For the left side, we add to both sides and shift the negative sign to the other side to get x=-6 .

For the right side, we add to both sides and x=6 .

x=6, x=-6

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Example Question #12 : Absolute Value

Solve for .

$6+\left | x\right |=0$

Possible Answers:

$\pm6$

No possible answer

Correct answer:

No possible answer

Explanation:

Let's isolate the variable by subtracting both sides by . We have:

$\left | x\right |=-6$ This will be a contradicting expression. Absolute values always generate positive values. Therefore no possible answer exist.

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Example Question #334 : Operations And Properties

Solve for .

$\frac{6}{\left | x\right |}=9$

Possible Answers:

$\frac{2}{3}$

$\pm \frac{3}{2}$

$\pm \frac{2}{3}$

$\frac{3}{2}$

Correct answer:

$\pm \frac{2}{3}$

Explanation:

When taking absolute values, we need to consider both positive and negative values. Let's multiply both sides by $\left | x\right |$ to get rid of the fraction. So, we have two equations. 9x=6, -(9x)=6

For the left equation, when we divide both sides by , $x=\frac{2}{3}$ .

For the right equation, we distribute the negative sign to get -9x=6 . When we divide both sides by , $x=-\frac{2}{3}$ .

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

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Example Question #335 : Operations And Properties

Solve for .

$\frac{\left | 4x\right |}{3}=8$

Possible Answers:

$\pm \frac{1}{6}$

$\frac{1}{6}$

$\pm6$

Correct answer:

$\pm6$

Explanation:

When taking absolute values, we need to consider both positive and negative values. Let's multiply each side by to get rid of the fraction. So, we have two equations. 4x=24, -(4x)=24

For the left equation, when we divide both sides by , x=6 .

For the right equation, we distribute the negative sign to get -4x=24 . When we divide both sides by , x=-6 .

x=6, x=-6

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Example Question #13 : Absolute Value

Solve for .

$\left | x\right |=17$

Possible Answers:

0, 17

$\pm17$

Correct answer:

$\pm17$

Explanation:

When taking absolute values, we need to consider both positive and negative values. So, we have two answers. x=17, x=-17

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Pre-Algebra : Absolute Value

Study concepts, example questions & explanations for Pre-Algebra

All Pre-Algebra Resources

Example Questions

Example Question #323 : Operations And Properties

Example Question #324 : Operations And Properties

Example Question #325 : Operations And Properties

Example Question #14 : Absolute Value

Example Question #11 : Absolute Value

Example Question #332 : Operations And Properties

Example Question #12 : Absolute Value

Example Question #334 : Operations And Properties

Example Question #335 : Operations And Properties

Example Question #13 : Absolute Value

All Pre-Algebra Resources

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