Enter an equation or problem

Camera input is not recognized!

Solution - Absolute value equations

Exact form:

x = 15, \frac{1}{7}

x=15 , \frac{1}{7}

Decimal form:

x = 15, 0.143

x=15 , 0.143

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
$| 11 x - 9 | = | 10 x + 6 |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| 11 x - 9 \| = \| 10 x + 6 \|$
$x = + y$	$(11 x - 9) = (10 x + 6)$
$x = - y$	$(11 x - 9) = - (10 x + 6)$
$+ x = y$	$(11 x - 9) = (10 x + 6)$
$- x = y$	$- (11 x - 9) = (10 x + 6)$

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:

$\| x \| = \| y \|$	$\| 11 x - 9 \| = \| 10 x + 6 \|$
$x = + y$ , $+ x = y$	$(11 x - 9) = (10 x + 6)$
$x = - y$ , $- x = y$	$(11 x - 9) = - (10 x + 6)$

2. Solve the two equations for x

7 additional steps

$(11 x - 9) = (10 x + 6)$

Subtract from both sides:

$(11 x - 9) - 10 x = (10 x + 6) - 10 x$

Group like terms:

$(11 x - 10 x) - 9 = (10 x + 6) - 10 x$

Simplify the arithmetic:

$x - 9 = (10 x + 6) - 10 x$

Group like terms:

$x - 9 = (10 x - 10 x) + 6$

Simplify the arithmetic:

$x - 9 = 6$

Add to both sides:

$(x - 9) + 9 = 6 + 9$

Simplify the arithmetic:

$x = 6 + 9$

Simplify the arithmetic:

$x = 15$

12 additional steps

$(11 x - 9) = - (10 x + 6)$

Expand the parentheses:

$(11 x - 9) = - 10 x - 6$

Add to both sides:

$(11 x - 9) + 10 x = (- 10 x - 6) + 10 x$

Group like terms:

$(11 x + 10 x) - 9 = (- 10 x - 6) + 10 x$

Simplify the arithmetic:

$21 x - 9 = (- 10 x - 6) + 10 x$

Group like terms:

$21 x - 9 = (- 10 x + 10 x) - 6$

Simplify the arithmetic:

$21 x - 9 = - 6$

Add to both sides:

$(21 x - 9) + 9 = - 6 + 9$

Simplify the arithmetic:

$21 x = - 6 + 9$

Simplify the arithmetic:

$21 x = 3$

Divide both sides by :

$\frac{(21 x)}{21} = \frac{3}{21}$

Simplify the fraction:

$x = \frac{3}{21}$

Find the greatest common factor of the numerator and denominator:

$x = \frac{(1 \cdot 3)}{(7 \cdot 3)}$

Factor out and cancel the greatest common factor:

$x = \frac{1}{7}$

3. List the solutions

$x = 15, \frac{1}{7}$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | 11 x - 9 |$
$y = | 10 x + 6 |$
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 x

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 x

3. List the solutions

4. Graph

Why learn this

Terms and topics

Related links

Latest Related Drills Solved