Math
Solve for both systems 3y+z=1 and 12y + 5z = 5
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Asked by
AnneS2
Last updated by
anonymous
3y+z=1
3y-3y+z=1-3y
z=1-3y
12y + 5z = 5
12y + 5(1-3y) = 5
12y + 5(1) (5)(-3y) = 5
12y+5-15y=5
5-3y=5
5-5-3y=5-5
-3y=0
-3y/-3=0/-3
y=0
z=1-3y
z=1-3(0)
z=1-0=1