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

Related questions
Impact of this question
1447 views around the world
You can reuse this answer
Creative Commons License