# What do you do when you have absolute values on both sides of the equations?

Jul 23, 2018

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#### Explanation:

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When we have absolute values on both sides of the equations,

we must consider both possibilities for acceptable solutions - positive and negative absolute value expressions.

We will look at an example first to understand:

Example-1

Solve for color(red)(x:

color(blue)(|2x-1|=|4x+9|

Both sides of the equation contain absolute values.

Find solutions as shown below:

color(red)((2x-1)=-(4x+9) .. Exp.1

color(blue)(OR

color(red)((2x-1)=(4x+9) ...Exp.2

color(green)(Case.1:

Consider ... Exp.1 first and solve for color(red)(x

color(red)((2x-1)=-(4x+9)

$\Rightarrow 2 x - 1 = - 4 x - 9$

Add color(red)(4x to both sides of the equation.

$\Rightarrow 2 x - 1 + 4 x = - 4 x - 9 + 4 x$

$\Rightarrow 2 x - 1 + 4 x = - \cancel{4 x} - 9 + \cancel{4 x}$

$\Rightarrow 6 x - 1 = - 9$

Add color(re)(1 to both sides of the equation.

$\Rightarrow 6 x - 1 + 1 = - 9 + 1$

$\Rightarrow 6 x - \cancel{1} + \cancel{1} = - 9 + 1$

$\Rightarrow 6 x = - 8$

Divide both sides by color(red)(2

$\Rightarrow \frac{6 x}{2} = - \frac{8}{2}$

$\Rightarrow 3 x = - 4$

color(blue)(rArr x = (-4/3) ... Sol.1

color(green)(Case.2:

Consider ... Exp.2 next and solve for color(red)(x

color(red)((2x-1)=(4x+9)

$\Rightarrow 2 x - 1 = 4 x + 9$

Subtract color(red)((4x) from both sides of the equation.

$\Rightarrow 2 x - 1 - 4 x = 4 x + 9 - 4 x$

$\Rightarrow 2 x - 1 - 4 x = \cancel{4 x} + 9 - \cancel{4 x}$

$\Rightarrow - 2 x - 1 = 9$

Add color(red)(1 to both sdies of the equation.

$\Rightarrow - 2 x - 1 + 1 = 9 + 1$

$\Rightarrow - 2 x - \cancel{1} + \cancel{1} = 9 + 1$

$\Rightarrow - 2 x = 10$

Divide both sides of the equation by color(red)(2

$\Rightarrow \frac{- 2 x}{2} = \frac{10}{2}$

$\Rightarrow - x = 5$

color(blue)(rArr x=-5 ... Sol.2

Hence, there are two solutions for the absolute value equation:

color(blue)(rArr x = (-4/3) ... Sol.1

color(blue)(rArr x=-5 ... Sol.2

If you so wish, you can substitute these values of color(red)(x in both color(green)(Case.1 and color(green)(Case.2 to verify the accuracy.

We will work on Example.2 in my next answer.

Hope it helps.

Jul 23, 2018

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Example.2 is given here.

#### Explanation:

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This is a continuation of my solution given earlier.

We worked on Example.1 in that solution.

Please refer to that solution first, before reading this solution.

Let us consider a second example:

Example.2

Solve for color(red)(x:

color(red)(5|x+3|-4=8|x+3|-4

Subtract color(blue)(8|x+3| and add color(blue)(4 on both sides:

$\Rightarrow 5 | x + 3 | - 4 - 8 | x + 3 | + 4 = 8 | x + 3 | - 4 - 8 | x + 3 | + 4$

$\Rightarrow 5 | x + 3 | - \cancel{4} - 8 | x + 3 | + \cancel{4} = \cancel{8 | x + 3 |} - 4 - \cancel{8 | x + 3 |} + 4$

$\Rightarrow 5 | x + 3 | - 8 | x + 3 | = - 4 + 4$

$\Rightarrow - 3 | x + 3 | = 0$

Divide both sides by color(red)((-3)

rArr [(-3)(|x+3|)]/((-3))=0/((-3)

$\Rightarrow \frac{\cancel{- 3} \left(| x + 3 |\right)}{\cancel{- 3}} = 0$

$\Rightarrow | x + 3 | = 0$

$\Rightarrow x + 3 = 0$

Subtract color(red)(3 from both sides

$\Rightarrow x + 3 - 3 = 0 - 3$

$\Rightarrow x + \cancel{3} - \cancel{3} = - 3$

$\Rightarrow x = - 3$

Hence, we conclude that

color(blue)(x=-3 is the ONLY Solution for this example.

Hope it helps.