Math
Solve for both systems 2z+4x=32 and 6z+5x=96

Asked by
AnneS2
Last updated by
anonymous
2z+4x=32
2z+4x-4x=32-4x
2z=32-4x
2z/2=32-4x/2
z=16-2x
6z+5x=96
6(16-2x)+5x=96
6(16) (6)(-2x)+5x=96
96-12x+5x=96
96-7x=96
96-7x-96=96-96
-7x=0
-7x/-7=0/-7
x=0
z=16-2x
z=16-2(0)
z=16