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]}