Math

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

Asked by AnneS2 on 22 Jul 05:01

Last updated by anonymous on 24 Jul 11:28

1 Answers

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Answered by CherieS on 24 Jul 11:28

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