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Solution - Absolute value equations

Exact form:

b = \frac{1}{123}, - \frac{1}{121}

b=\frac{1}{123} , -\frac{1}{121}

Decimal form:

b = 0.008, - 0.008

b=0.008 , -0.008

See steps

Step-by-step explanation

1. Rewrite the equation without absolute value bars

Use the rules:
$| x | = | y |$ → $x = \pm y$ and $| x | = | y |$ → $\pm x = y$
to write all four options of the equation
$| 122 b | = - | b - 1 |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| 122 b \| = - \| b - 1 \|$
$x = + y$	$(122 b) = - (b - 1)$
$x = - y$	$(122 b) = - (- (b - 1))$
$+ x = y$	$(122 b) = - (b - 1)$
$- x = y$	$- (122 b) = - (b - 1)$

When simplified, equations $x = + y$ and $+ x = y$ are the same and equations $x = - y$ and $- x = y$ are the same, so we end up with only 2 equations:

$\| x \| = \| y \|$	$\| 122 b \| = - \| b - 1 \|$
$x = + y$ , $+ x = y$	$(122 b) = - (b - 1)$
$x = - y$ , $- x = y$	$(122 b) = - (- (b - 1))$

2. Solve the two equations for b

6 additional steps

$122 b = - (b - 1)$

Expand the parentheses:

$122 b = - b + 1$

Add to both sides:

$(122 b) + b = (- b + 1) + b$

Simplify the arithmetic:

$123 b = (- b + 1) + b$

Group like terms:

$123 b = (- b + b) + 1$

Simplify the arithmetic:

$123 b = 1$

Divide both sides by :

$\frac{(123 b)}{123} = \frac{1}{123}$

Simplify the fraction:

$b = \frac{1}{123}$

6 additional steps

$122 b = - (- (b - 1))$

NT_MSLUS_MAINSTEP_RESOLVE_DOUBLE_MINUS:

$122 b = b - 1$

Subtract from both sides:

$(122 b) - b = (b - 1) - b$

Simplify the arithmetic:

$121 b = (b - 1) - b$

Group like terms:

$121 b = (b - b) - 1$

Simplify the arithmetic:

$121 b = - 1$

Divide both sides by :

$\frac{(121 b)}{121} = \frac{- 1}{121}$

Simplify the fraction:

$b = \frac{- 1}{121}$

3. List the solutions

$b = \frac{1}{123}, - \frac{1}{121}$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | 122 b |$
$y = - | b - 1 |$
The equation is true where the two lines cross.

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Why learn this

We encounter absolute values almost daily. For example: If you walk 3 miles to school, do you also walk minus 3 miles when you go back home? The answer is no because distances use absolute value. The absolute value of the distance between home and school is 3 miles, there or back.
In short, absolute values help us deal with concepts like distance, ranges of possible values, and deviation from a set value.

Solution - Absolute value equations

Step-by-step explanation

1. Rewrite the equation without absolute value bars

2. Solve the two equations for b

3. List the solutions

4. Graph

Why learn this

Terms and topics

Related links

Solution - Absolute value equations

Step-by-step explanation

1. Rewrite the equation without absolute value bars

2. Solve the two equations for b

3. List the solutions

4. Graph

Why learn this

Terms and topics

Related links

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