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Solution - Absolute value equations

Exact form:

p = 6, \frac{2}{5}

p=6 , \frac{2}{5}

Decimal form:

p = 6, 0.4

p=6 , 0.4

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
$| 8 p - 6 | = | 7 p |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| 8 p - 6 \| = \| 7 p \|$
$x = + y$	$(8 p - 6) = (7 p)$
$x = - y$	$(8 p - 6) = - (7 p)$
$+ x = y$	$(8 p - 6) = (7 p)$
$- x = y$	$- (8 p - 6) = (7 p)$

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:

$\| x \| = \| y \|$	$\| 8 p - 6 \| = \| 7 p \|$
$x = + y$ , $+ x = y$	$(8 p - 6) = (7 p)$
$x = - y$ , $- x = y$	$(8 p - 6) = - (7 p)$

2. Solve the two equations for p

6 additional steps

$(8 p - 6) = 7 p$

Subtract from both sides:

$(8 p - 6) - 7 p = (7 p) - 7 p$

Group like terms:

$(8 p - 7 p) - 6 = (7 p) - 7 p$

Simplify the arithmetic:

$p - 6 = (7 p) - 7 p$

Simplify the arithmetic:

$p - 6 = 0$

Add to both sides:

$(p - 6) + 6 = 0 + 6$

Simplify the arithmetic:

$p = 0 + 6$

Simplify the arithmetic:

$p = 6$

9 additional steps

$(8 p - 6) = - 7 p$

Add to both sides:

$(8 p - 6) + 6 = (- 7 p) + 6$

Simplify the arithmetic:

$8 p = (- 7 p) + 6$

Add to both sides:

$(8 p) + 7 p = ((- 7 p) + 6) + 7 p$

Simplify the arithmetic:

$15 p = ((- 7 p) + 6) + 7 p$

Group like terms:

$15 p = (- 7 p + 7 p) + 6$

Simplify the arithmetic:

$15 p = 6$

Divide both sides by :

$\frac{(15 p)}{15} = \frac{6}{15}$

Simplify the fraction:

$p = \frac{6}{15}$

Find the greatest common factor of the numerator and denominator:

$p = \frac{(2 \cdot 3)}{(5 \cdot 3)}$

Factor out and cancel the greatest common factor:

$p = \frac{2}{5}$

3. List the solutions

$p = 6, \frac{2}{5}$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | 8 p - 6 |$
$y = | 7 p |$
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 p

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 p

3. List the solutions

4. Graph

Why learn this

Terms and topics

Related links

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