Enter an equation or problem

Camera input is not recognized!

Solution - Absolute value equations

Exact form:

d = 5, \frac{5}{11}

d=5 , \frac{5}{11}

Decimal form:

d = 5, 0.455

d=5 , 0.455

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
$| 5 d | = | 6 d - 5 |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| 5 d \| = \| 6 d - 5 \|$
$x = + y$	$(5 d) = (6 d - 5)$
$x = - y$	$(5 d) = - (6 d - 5)$
$+ x = y$	$(5 d) = (6 d - 5)$
$- x = y$	$- (5 d) = (6 d - 5)$

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:

$\| x \| = \| y \|$	$\| 5 d \| = \| 6 d - 5 \|$
$x = + y$ , $+ x = y$	$(5 d) = (6 d - 5)$
$x = - y$ , $- x = y$	$(5 d) = - (6 d - 5)$

2. Solve the two equations for d

6 additional steps

$5 d = (6 d - 5)$

Subtract from both sides:

$(5 d) - 6 d = (6 d - 5) - 6 d$

Simplify the arithmetic:

$- d = (6 d - 5) - 6 d$

Group like terms:

$- d = (6 d - 6 d) - 5$

Simplify the arithmetic:

$- d = - 5$

Multiply both sides by :

$- d \cdot - 1 = - 5 \cdot - 1$

Remove the one(s):

$d = - 5 \cdot - 1$

Simplify the arithmetic:

$d = 5$

6 additional steps

$5 d = - (6 d - 5)$

Expand the parentheses:

$5 d = - 6 d + 5$

Add to both sides:

$(5 d) + 6 d = (- 6 d + 5) + 6 d$

Simplify the arithmetic:

$11 d = (- 6 d + 5) + 6 d$

Group like terms:

$11 d = (- 6 d + 6 d) + 5$

Simplify the arithmetic:

$11 d = 5$

Divide both sides by :

$\frac{(11 d)}{11} = \frac{5}{11}$

Simplify the fraction:

$d = \frac{5}{11}$

3. List the solutions

$d = 5, \frac{5}{11}$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | 5 d |$
$y = | 6 d - 5 |$
The equation is true where the two lines cross.

How did we do?

Please leave us feedback.

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 d

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 d

3. List the solutions

4. Graph

Why learn this

Terms and topics

Related links

Latest Related Drills Solved