Split equal absolute values into the two signed linear equations.
Notes and key methods
For real U,V, \(|U|=|V|\) is equivalent to \(\displaystyle U=V\) or \(\displaystyle U=-V\). Solve both equations and remove duplicate roots. Squaring is also valid for this equality, but can hide the useful linear factors.
When solving an inequality, split at the zeros of the expressions inside the moduli or compare squared nonnegative sides. Test an interval, rather than carrying an equality argument over without checking the sign.
Check the conditions
Both expressions are real.
An inequality requires sign analysis after finding its boundaries.
Three worked examples
Example 1 · Two modulus expressions
Write \(\displaystyle U=\)\(3 x + 5\) and \(\displaystyle V=\)\(5 x + 11\). Consider \(\left|{3 x + 5}\right| = \left|{5 x + 11}\right|\).
Solve the modulus equation completely.
Solve \(\displaystyle |U|\le |V|\) for these two expressions.
Find the distance between the two solutions.
Find the midpoint of the two solutions.
Show the reasoning
(a)
\(3 x + 5 = 5 x + 11\) and \(3 x + 5 = - 5 x - 11\) give the displayed two roots.
Answer: \(\left\{-3, -2\right\}\)
(b)
\(\left(3 x + 5\right)^{2} - \left(5 x + 11\right)^{2} \leq 0\) has negative leading coefficient, so it holds outside and at the two roots.