How do you solve #|5x | = | 3x - 8|#?

1 Answer
Jun 18, 2018

Consider the various cases of #x# ranges of the moduli and solve within them

Explanation:

Deduce piecewise domains of the equation

The modulus on the LHS changes behaviour at #x=0#; the modulus on the RHS at #3x-8=0rArrx=8/3#.

Take the cases for the modulus on the LHS first:
If #x>=0#, then #5x=|3x-8|#
If #x<0#, then #-5x=|3x-8|#

Now for both of those, consider the two cases for the modulus on the RHS:
If #x>=0# and #5x=|3x-8|#, then #x<8/3# means that #5x=-(3x-8)=-3x+8#.
If #x>=0# and #5x=|3x-8|#, then #x>=8/3# means that #5x=3x-8#.
If #x<0# and #-5x=|3x-8|#, then #x# is always less than 8/3 from the first condition already, so #-5x=-(3x-8)=-3x+8rArr 5x=3x-8#.
The final case never happens because the two conditions have zero overlap: #x<0# and #x>=8/3#.

So we have an equation defined in three pieces according to ranges of #x#:
If #x<0#, then #5x=3x-8#
If #0<=x<8/3#, then #5x=-3x+8#
If #8/3<=x#, then #5x=3x-8#
So it changes one way, then the other, as #x# increases.

Solve the equation on each domain

We have two different function forms - one of which applies to two of the domains.
#5x=3x-8rArr2x=-8rArrx=-4#
#5x=-3x+8rArr8x=8rArrx=1#

Matching these to the domains above, we see that the correct equation has been solved for the answer that lies in #x<0#, and ditto for the answer that lies in #0<=x<8/3#. There is no answer in the third domain #8/3<=x#.

So we have two valid answers to the equation:
#x=-4, 1#

Sanity check the answer by overplotting the two sides of the graph, seeing that they cross at these values:
graph{(y-|5x|)(y-|3x-8|)=0 [-10, 10, -6.4, 33.6]}