Can you show how you solve this math problem please

3 - 4(z+1) = 3z - (2+z) is the question. But I'm stumped. So would you kindly show me how you do it? Your supposed to answer "as a simple fraction." Thanks.

Well all we have to do here is solve for z.
Start by simplifying the equation. For the left side:
3 - 4(z+1)
you have to multiply -4 through the parentheses. this yields:
3 - 4z - 4
which we can simplify to -4z - 1
For the right side:
3z - (2 + z)
just multiply the negative through the parentheses (in other words, negate it). This yields:
3z - 2 - z or, simplified, is 2z - 2
Now our equation is: -4z - 1 = 2z - 2
Transpose all the terms with z to one side and everything else to the other.
-1 + 2 = 2z + 4z (I transposed to the right to keep z positive)
Simplify both sides.
1 = 6z
Divide both sides by 6 to isolate z.
1/6 = z OR z = 1/6
This is already in the lowest term, so that's your final answer.

Start by getting rid of the parentheses. Multiply the 4 on the left and the implied -1 on the right to get,

3 - 4z - 4 = 3z - 2 -z ======> - 4z - 1 = 2z - 2

Now get like terms to seperate sides (i.e z's on the left numbers on the right)

2-1= 2z + 4z ====> 1= 6z

divide both sides by 6 to get the z by itself.

z = 1/6

1) 3 - 4(z+1) = 3z - (2+z)

2) 3 - 4z - 4 = 3z - 2 - z

3) -1 - 4z = 2z - 2

4) -6z = -1

5) z = 1/6

BTW, contributor Dave is obviously gaming the system.

3 - 4(z+1) = 3z - (2+z)

3 - 4z - 4 = 3z - 2 - z

- 4z - 3z + z = - 3 + 4 - 2

- 6z = - 1

z = (-1)/(-6)

z = 1/6

3-4z-4=3z-2-z

-4z-4=2z-2
-2=2z
-1=z

3 - 4z - 4 = 3 z - 2 - z

take z term one side

-4z- 3z+ z = -3-2+4

-6z= -1

z=1/6