what is 8z+20= - |2z+4|

Absolute value equalitiy entered
      8z+20 = -|2z+4|

Absolute value term is moved to the left hand side.
Two terms are moved / added to the right hand side.

      |2z+4| = -8z-20 Clear the absolute-value bars by splitting the equation into its two cases, one for the Positive case and the other for the Negative case.

The Absolute Value term is |2z+4|

For the Negative case we'll use -(2z+4)

For the Positive case we'll use (2z+4)    -(2z+4) = -8z-20

     Multiply
      -2z-4 = -8z-20

     Rearrange and Add up
      6z = -16

     Divide both sides by 6
      z = -(8/3)  (2z+4) = -8z-20

     Rearrange and Add up
      10z = -24

     Divide both sides by 10
      z = -(12/5)

     Which is the solution for the Positive Case

z=-8/3
z=-12/5

dalendrk dalendrk · Answer 2 · 2016-08-31T23:35:27+02:00

[tex]|a|= \left\{\begin{array}{ccc}a&for\ a\geq0\\-a&for\ a \ \textless \ 0\end{array}\right\\\\2z+4\geq0\ \ \ \ |subtract\ 4\ from\ both\ sides\\\\2z\geq-4\ \ \ \ \ |divide\ both\ sides\ by\ 2\\\\z\geq-2\\\\|2z+4|= \left\{\begin{array}{ccc}2z+4&for\ z\geq-2\\-(2z+4)&for\ z \ \textless \ -2\end{array}\right\\\\1^o\ z \ \textless \ -2\to z\in(-\infty;-2) \\\\8z+20=-[-(2z+4)]\\8z+20=2z+4\ \ \ \ \ |subtract\ 20\ from\ both\ sides\\8z=2z-16\ \ \ \ \ |subtract\ 2z\ from\ both\ sides\\6z=-16\ \ \ \ \ |divide\ both\ sides\ by\ 6\\z=-2\dfrac{2}{3}\ \in\ 1^o[/tex]

[tex]2^o\ z\geq-2\to z\in[-2;\ \infty)\\\\8z+20=-(2z+4)\\8z+20=-2z-4\ \ \ \ |subtract\ 20\ from\ both\ sides\\8z=-2z-24\ \ \ \ |add\ 2z\ to\ both\ sides\\10z=-24\ \ \ \ \ \ |divide\ both\ sides\ by\ 10\\z=-2.4\ \not\in\ 2^o\\\\Answer:\boxed{z=-2\frac{2}{3}}[/tex]