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Given that x < 5, rewrite 8x + 5 - |x - 5| without using absolute value signs.

 Jun 13, 2021
 #1
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$8x+5-\sqrt{(x-5)^2}$

 Jun 13, 2021
 #2
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The above answer works for all $x$.  If we know that $x<5$, we have
$8x+5-|x-5|= 8x+5 - (5-x) = 8x+5-5+x= 9x$.
 

Bginner  Jun 13, 2021
 #3
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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 \(9x\).

 Jun 14, 2021

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