Answer
$(3x-5)^\circ=40^\circ$
$(6x-40)^\circ=50^\circ$
Work Step by Step
If two angles are complementary, the sum of the two angles is a right angle ($90^\circ$):
$(3x-5)^\circ+(6x-40)^\circ=(9x-45)^\circ=90^\circ$
$9x-45=90$
$9x=135$
$x=15$
$(3x-5)^\circ=40^\circ$
$(6x-40)^\circ=50^\circ$
