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

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Answer 1 · 2018-06-13T16:23:53

$z \leq - 3$ $z<=-3$

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

$z + 3 z + 8 \leq - 4$ $z+3z+8 le -4$

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

$4 z + 8 \leq - 4$ $4z+8 le -4$

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

$4 z + 8 - 8 \leq - 4 - 8$ $4z+8-8 le -4-8$

$4 z \leq - 12$ $4z le -12$

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

$\frac{4 z}{4} \leq \frac{- 12}{4}$ $(4z)/4 le (-12)/4$

$z \leq - 3$ $z le -3$

And that's your answer!

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