## Precalculus (6th Edition)

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