+118696 Thanks Chris,
This is what I would do with this question.
36≤1−5z>−21$subtract1throughout$35≤−5z>−22$dividethroughoutby−5butyouwillhavetoturnthesignsaround$−7≥z<−4.4$nowIwillseparatethembecausethisreallydoesn′tmakemuchsense.$−7≥zandz<−4.4z≤−7andz<−4.4soZ≤−7
This original line 36≤1−5z>−21 IS NOT normal or good presentation.
It should have been presented as
36≤1−5zand1−5z>−21
+130462 36 ≤ 1 - 5z > -21 suubtract 1 from everything
35 ≤ -5z > -22 divide everything by -5 and don't forget that we have to "reverse" the signs when dividing by a negative!!
-7 ≥ z < 22/5
Note that we must use the most restrictive interval, here ..... this is ... -7 ≥ z
+143 would you not add 5 to cancel the 5out and then add 5 to the -22 to make it -17
+130462 No....remember.....it's -5z .....not just -5...... we need to subtract 1 in the middle at the start to get just -5z remaining.....
+118696 Thanks Chris,
This is what I would do with this question.
36≤1−5z>−21$subtract1throughout$35≤−5z>−22$dividethroughoutby−5butyouwillhavetoturnthesignsaround$−7≥z<−4.4$nowIwillseparatethembecausethisreallydoesn′tmakemuchsense.$−7≥zandz<−4.4z≤−7andz<−4.4soZ≤−7
This original line 36≤1−5z>−21 IS NOT normal or good presentation.
It should have been presented as
36≤1−5zand1−5z>−21
