How do you solve #2(3z-5)+4(z+6)>=2(3z+2) +3z-15#?

1 Answer
kat
Dec 5, 2017

#z ≥ -3#

Explanation:

STEP ONE: Distribute the equation
The ones in bold/italics are distributed to the one below it

2(3z-5) + 4(z+6) ≥ 2(3z+2) +3z-15

6z-10 +4z+24 ≥ 6z+4 +3z -15

STEP TWO: Add all the number together that have the same variable (so #z# variables go together and number that do not have a variable go together

#6z-10+4z+24 ≥ 6z+4+3z -15#

#10z+14 ≥ 9z -11#

STEP THREE: Switch the #z# variable numbers to the side that you want the #z# variable to be on (I'm choosing the left side) and switch the no variable numbers to the side you want them to be on (I'm choosing the right side)

#10z+14 ≥ 9z -11#
#10z-9z ≥ -11-14#
#z ≥ -3#

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