Question

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

Answer 1

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]}