SOLUTION: 1) 4(a-3)=3-(a+5) 2) 3z=2(z-3)+9 3) 8a-(a-7)=21 I am in an adult ed program getting my high school diploma. These three problems are the only ones that I am struggling re

Algebra ->  Equations -> SOLUTION: 1) 4(a-3)=3-(a+5) 2) 3z=2(z-3)+9 3) 8a-(a-7)=21 I am in an adult ed program getting my high school diploma. These three problems are the only ones that I am struggling re      Log On


   

Question 183802: 1) 4(a-3)=3-(a+5)
2) 3z=2(z-3)+9
3) 8a-(a-7)=21
I am in an adult ed program getting my high school diploma. These three problems are the only ones that I am struggling real hard with. We are given packets to work--and I have grasped most of the problems. These three are the ones that are not computing for me. HELP!!
Found 2 solutions by stanbon, halbren:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
1) 4(a-3)=3-(a+5)
4a - 12 = 3 - a - 5
5a -12 = -2
5a = 10
a = 2
========================
2) 3z=2(z-3)+9
3z = 2z - 6 + 9
z = 3
=======================
3) 8a-(a-7)=21
8a - a + 7 = 21
7a = 14
a = 2
===========================
Cheers,
Stan H.

Answer by halbren(10) About Me  (Show Source):
You can put this solution on YOUR website!
Problem 1
First eliminate the parentheses by distribution
4a-%284%29%283%29=+3-a-5
remember that a - sign outside a ( stands for -1 so you need to multiply everything inside the parentheses by -1
simplify
4a-12=-a-2
Add 12 to both sides
4a-12%2B12=-a-2%2B12
4a=-a%2B10
Add a to both sides
4a%2Ba=-a%2Ba%2B10
5a=10
Divide both sides by 5
5a%2F5=10%2F5
a=2
Problem 2 is similar
3z=2z-2%2A3%2B9
3z=2z-6%2B9
3z=2z%2B3
Subtract 2z from both sides
3z-2z=2z-2z%2B3
z=3
Problem 3 has the - sign like problem 1. You should be able to follow the same steps