Absolute Value Equations

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Question 1 of 5

1. Question
Illustrate $| x | = 3$ $|x|=3$ using a number line.
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Absolute value describes the distance of a number on the number line from $0$ $0$ without considering which direction from zero the number lies.

First, find the possible values of $x$ $x$

$| x | = 3$ $|x|=3$ means the distance of $x$ $x$ from $0$ $0$ is $3$ $3$ units. Therefore, $x = 3$ $x=3$ or $x = - 3$ $x=-3$

Next, plot $3$ $3$ and $- 3$ $-3$ on the number line.

$x = 3$ $x=3$

$x = - 3$ $x=-3$
Question 2 of 5

2. Question
Solve for $x$ $x$

$| x - 5 | = 11$ $|x-5|=11$
- 1. $x = 13$ $x=13$ or $- 3$ $-3$
- 2. $x = 16$ $x=16$ or $- 6$ $-6$
- 3. $x = - 13$ $x=-13$ or $3$ $3$
- 4. $x = 6$ $x=6$ or $- 16$ $-16$
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Absolute value describes the distance of a number on the number line from $0$ $0$ without considering which direction from zero the number lies.

Since we are finding the absolute value of $x$ $x$ , form a positive and negative equation and solve for $x$ $x$ on both equations

Positive:

$x - 5$ $x-5$ $=$ $=$ $11$ $11$

$x - 5$ $x-5$ $+ 5$ $+5$ $=$ $=$ $11$ $11$ $+ 5$ $+5$ Add $5$ $5$ to both sides

$x$ $x$ $=$ $=$ $16$ $16$

Negative:

$x - 5$ $x-5$ $=$ $=$ $- 11$ $-11$

$x - 5$ $x-5$ $+ 5$ $+5$ $=$ $=$ $- 11$ $-11$ $+ 5$ $+5$ Add $5$ $5$ to both sides

$x$ $x$ $=$ $=$ $- 6$ $-6$

To check if the values of $x$ $x$ are correct, substitute the values to the equation and check if its absolute value is $11$ $11$

$x = 16$ $x=16$

$| x - 5 |$ $|x-5|$ $=$ $=$ $11$ $11$

$| 16 - 5 |$ $|16-5|$ $=$ $=$ $11$ $11$ $x = 16$ $x=16$

$|11|$ $=$ $11$

$11$ $=$ $11$ The absolute value of $11$ is $11$

$x=-6$

$|x-5|$ $=$ $11$

$|-6-5|$ $=$ $11$ $x=16$

$|-11|$ $=$ $11$

$11$ $=$ $11$ The absolute value of $-11$ is $11$

Therefore, the absolute values of $x$ is $16$ or $-6$

$x=16$ or $-6$

Question 3 of 5

Solve for

$x$

$4|2x-2|-14=10$

1. $x=-4$ or $4$
2. $x=1$ or $-2$
3. $x=4$ or $-2$
4. $x=-4$ or $2$

Question 4 of 5

Solve for

$x$

$-4|1-3/4 x|=-16$

1. $x=-6$ or $4 2/3$
2. $x=6$ or $-4 2/3$
3. $x=4$ or $-6 2/3$
4. $x=-4$ or $6 2/3$

Question 5 of 5

Solve for

$y$

$-7|2y-2|+5=-37$

1. $y=2$ or $-2$
2. $y=-4$ or $1$
3. $y=3$ or $-2$
4. $y=4$ or $-2$

$4\|2x-2\|-14$	$=$	$10$
$4\|2x-2\|-14$ $+14$	$=$	$10$ $+14$	Add $14$ to both sides
$4\|2x-2\|$	$=$	$24$
$4\|2x-2\|$ $div4$	$=$	$24$ $div4$	Divide both sides by $4$
$\|2x-2\|$	$=$	$6$

$2x-2$	$=$	$6$
$2x-2$ $+2$	$=$	$6$ $+2$	Add $2$ to both sides
$2x$	$=$	$8$
$2x$ $div2$	$=$	$8$ $div2$	Divide both sides by $2$
$x$	$=$	$4$

$2x-2$	$=$	$-6$
$2x-2$ $+2$	$=$	$-6$ $+2$	Add $2$ to both sides
$2x$	$=$	$-4$
$2x$ $div2$	$=$	$-4$ $div2$	Divide both sides by $2$
$x$	$=$	$-2$

$1-3/4 x$	$=$	$4$

$1-3/4 x$ $-1$	$=$	$4$ $-1$	Subtract $1$ from both sides

$-3/4 x$	$=$	$3$

$-3/4 x$ $times(-4)$	$=$	$3$ $times(-4)$	Multiply both sides by $-4$

$3x$	$=$	$-12$
$3x$ $div3$	$=$	$-12$ $div3$	Divide both sides by $3$
$x$	$=$	$-4$

$1-3/4 x$	$=$	$-4$

$1-3/4 x$ $-1$	$=$	$-4$ $-1$	Subtract $1$ from both sides

$-3/4 x$	$=$	$-5$

$-3/4 x$ $times(-4)$	$=$	$-5$ $times(-4)$	Multiply both sides by $-4$

$3x$	$=$	$20$
$3x$ $div3$	$=$	$20$ $div3$	Divide both sides by $3$

$x$	$=$	$20/3$
$x$	$=$	$6 2/3$	Convert to a mixed number

Absolute Value Equations

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3. Question

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Absolute Value Standard Form

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Absolute Value Standard Form

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Absolute Value Standard Form

Quizzes

$x - 5$ $x-5$	$=$ $=$	$11$ $11$
$x - 5$ $x-5$ $+ 5$ $+5$	$=$ $=$	$11$ $11$ $+ 5$ $+5$	Add $5$ $5$ to both sides
$x$ $x$	$=$ $=$	$16$ $16$

$x - 5$ $x-5$	$=$ $=$	$- 11$ $-11$
$x - 5$ $x-5$ $+ 5$ $+5$	$=$ $=$	$- 11$ $-11$ $+ 5$ $+5$	Add $5$ $5$ to both sides
$x$ $x$	$=$ $=$	$- 6$ $-6$

$\| x - 5 \|$ $\|x-5\|$	$=$ $=$	$11$ $11$
$\| 16 - 5 \|$ $\|16-5\|$	$=$ $=$	$11$ $11$	$x = 16$ $x=16$
$\|11\|$	$=$	$11$
$11$	$=$	$11$	The absolute value of $11$ is $11$

$\|x-5\|$	$=$	$11$
$\|-6-5\|$	$=$	$11$	$x=16$
$\|-11\|$	$=$	$11$
$11$	$=$	$11$	The absolute value of $-11$ is $11$

$-4\|1-3/4 x\|$	$=$	$-16$

$-4\|1-3/4 x\|$ $div(-4)$	$=$	$-16$ $div(-4)$	Divide both sides by $-4$

$\|1-3/4 x\|$	$=$	$4$

$-7\|2y-2\|+5$	$=$	$-37$
$-7\|2y-2\|+5$ $-5$	$=$	$-37$ $-5$	Subtract $5$ from both sides
$-7\|2y-2\|$	$=$	$-42$
$-7\|2y-2\|$ $div(-7)$	$=$	$-42$ $div(-7)$	Divide both sides by $-7$
$\|2y-2\|$	$=$	$6$

$2y-2$	$=$	$6$
$2y-2$ $+2$	$=$	$6$ $+2$	Add $2$ to both sides
$2y$	$=$	$8$
$2y$ $div2$	$=$	$8$ $div2$	Divide both sides by $2$
$y$	$=$	$4$

$-7\|2y-2\|+5$	$=$	$-37$
$-7\|2(4)-2\|+5$	$=$	$-37$	$y=4$
$-7\|8-2\|+5$	$=$	$-37$
$-7\|6\|+5$	$=$	$-37$
$-7(6)+5$	$=$	$-37$	The absolute value of $6$ is $6$
$-42+5$	$=$	$-37$
$-37$	$=$	$-37$

$-7\|2y-2\|+5$	$=$	$-37$
$-7\|2(-2)-2\|+5$	$=$	$-37$	$y=-2$
$-7\|-4-2\|+5$	$=$	$-37$
$-7\|-6\|+5$	$=$	$-37$
$-7(6)+5$	$=$	$-37$	The absolute value of $-6$ is $6$
$-42+5$	$=$	$-37$
$-37$	$=$	$-37$