Given that x < 5, rewrite 8x + 5 - |x - 5| without using absolute value signs.
The above answer works for all $x$. If we know that $x<5$, we have $8x+5-|x-5|= 8x+5 - (5-x) = 8x+5-5+x= 9x$.
If \(x <5\), then \(\left |x-5 \right |<0\). \(\left |x-5 \right |\) would be \(5-x\). So \(8x+5-\left |x-5 \right |\) would be \(9x\).