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

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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.

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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.