Solve for each system 4z+4y=92 and 3z + 5y = 109

4z+4y=92

4z-4z+4y=92-4z

4y=92-4z

4y/4=92-4z/4

y=23-z

3z + 5y = 109

3z + 5(23-z) = 109

3z + 5(23) (5)(-z) = 109

3z+115-5z=109

115-2z=109

115-2z-115=109-115

-2z=-6

-2z/-2=-6/-2

z=3

y=23-z

y=23-3=20