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How do you solve #6( z + 2) \leq 13#?

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Sep 2, 2017

#6(z+2) ≤ 13# is another way of saying#(6 xx z)+ (6 xx 2)<=13# .

#6z + 12 <= 13#

#6z + 12 -12 <= 13 - 12#

#6z <= 1#

#(6z)/(6z) <= 1/(6z)#

#z = 0.1666666667#

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