Question

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

Answer 1

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Please read the explanation.

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)`

`rArr 2x-1=-4x-9`

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

`rArr 2x-1+4x=-4x-9+4x`

`rArr 2x-1+4x=-cancel (4x)-9+cancel(4x)`

`rArr 6x-1=-9`

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

`rArr 6x-1+1=-9+1`

`rArr 6x-cancel 1+cancel 1=-9+1`

`rArr 6x=-8`

Divide both sides by `color(red)(2`

`rArr (6x)/2=-8/2`

`rArr 3x=-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)`

`rArr 2x-1=4x+9`

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

`rArr 2x-1-4x=4x+9-4x`

`rArr 2x-1-4x=cancel(4x)+9-cancel(4x)`

`rArr -2x-1=9`

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

`rArr -2x-1+1=9+1`

`rArr -2x-cancel 1+cancel 1=9+1`

`rArr -2x=10`

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

`rArr (-2x)/2=10/2`

`rArr -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.

Answer 2

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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:

`rArr 5|x+3|-4-8|x+3|+4=8|x+3|-4-8|x+3|+4`

`rArr 5|x+3|-cancel 4-8|x+3|+cancel 4=cancel(8|x+3|)-4-cancel(8|x+3|)+4`

`rArr 5|x+3|-8|x+3|=-4+4`

`rArr -3|x+3|=0`

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

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

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

`rArr |x+3|=0`

`rArr x+3=0`

Subtract `color(red)(3` from both sides

`rArr x+3-3=0-3`

`rArr x+cancel 3-cancel 3=-3`

`rArr x=-3`

Hence, we conclude that

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

Hope it helps.