Modulus equations with a linear right side

For |u|=v, require v 0, then solve u=v or u=-v. Squaring forgets the sign of v, so it can introduce false candidates. Write the right-side condition first. Solve both branches and check each candidate in the original equation. An identity on one branch gives an interval, not a single root. For |x+1|=2x, require x 0. The sign equations give x=1,- 13; only 1 is valid. Check: |x-2|=x-2 holds for every x 2. Never divide by a factor that may be zero or discard a branch identity.