Need help with HOW to solve this algebra question!!?

Question

I'm trying to explain this to someone and I can't remember how to deal with the fractions!

z-2/3(2z+3)=1/2(1-z)
The know the answer is z=3 but cant remember how to get there. Thanks!

The reason I said the answer is -3 is because that is the answer the book is giving

Answer 1

z - 2/3(2z + 3) = 1/2(1-z)
Multiply each term of both sides by 6
6z - 4(2z+3) = 3(1-z)
6z -8z - 12 = 3 - 3z
-2z -12 = 3 - 3z
z = 15

Plug it in to original equation for a check and you will see that this is the answer.

Answer 2

It is like any equation. Get rid of the denominators by multiplying by the least common denominator. In this case, it is 6.
z-2/3(2z+3)=1/2(1-z)
So,
6z-4(2z+3)=3(1-z)
Simplify,
6z-8z-12=3-3z
-2z-12=3-3z
Move around
z=15

Answer 3

first thing you do is multiply 2/3 with 2z and with 3
you get z -4/3z -2 = 1/2(1-z)
then you multiply 1/2 with 1 and with z
you get z -4/3z -2 = 1/2 -1/2z
then you subtract 4/3z from z
you get -1/3z -2 = 1/2 -1/2z
then +1/2z +1/2z
you get 1/6z -2 = 1/2
then +2 +2
you get 1/6z = 2 1/2
then divide each side by 1/6
you get z = 15

Answer 4

See 6=x/4+2 right here 4 has the sign of branch it is going to the left and multiplies 6*4=x+2 i.e. 24=x+2 now 2 is with + it is going to the left and is subtracted 24-2=x hence 22=x or x=22 Ans.

Answer 5

The answer should be -9!!!!
I figured it out and resulted in -9.
z-2/3(2z+3)=1/2(1-z)
z-(4z+6)over3 equals to (1-z) over 2
(3z-4z+6) over 3 equals to (1-z) over 2
2(-z+6)=3(1-z)
-2z+3z+12-3=0
z+9=0
z=-9

Answer 6

multiply by 6 so you get

6z-4(2z+3)=3(1-z)
6z-8z-12=3-3z
-2z-12=3-3z
z-12=3
z=15

If you plug 3 in for z you get: 3-6=-1?

Answer 7

isnt it z=2?? u sure its 3? or is my algebra dwindling?!?!?