How do you solve #z+ 8+ 3z \leq - 4#?

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Answer 1 · 2018-06-13T16:23:53

#z<=-3#

First, we can change the order of addition on the left hand side to group all the variable #z# terms.

#z+3z+8 le -4#

Now, notice that #z+3z=4z#.

#4z+8 le -4#

In order to get #z# by itself, subtract #8# from both sides of the inequality.

#4z+8-8 le -4-8#

#4z le -12#

Now, divide both sides of the inequality by #4#.

#(4z)/4 le (-12)/4#

#z le -3#

And that's your answer!