SOLUTION: 41: 4(3+x)+10= -11(x-6)+1 43:3x+21=-2x+11 45:6x+4=2x whenever i experiance any of these types of questions im not sure how to solve for x when there is an x on both sides of

Algebra ->  Exponents-negative-and-fractional -> SOLUTION: 41: 4(3+x)+10= -11(x-6)+1 43:3x+21=-2x+11 45:6x+4=2x whenever i experiance any of these types of questions im not sure how to solve for x when there is an x on both sides of       Log On


   

Question 148138This question is from textbook algebra: the easy way
: 41: 4(3+x)+10= -11(x-6)+1
43:3x+21=-2x+11
45:6x+4=2x
whenever i experiance any of these types of questions im not sure how to solve for x when there is an x on both sides of the equation. Given that the book that im using does not go over how to do these types of equations im at lost of how to do it, any help trying to solve these equations is greatly appreciated by me, thank you.
-Brian. This question is from textbook algebra: the easy way

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Just follow the basic rules of algebra to get the variable (x) on one side by itself
41:
4(3+x)+10 = -11(x-6) + 1
:
Multiply what's inside the brackets:
12 + 4x + 10 = -11x + 66 + 1
:
4x + 22 = -11x + 67
:
Add 11x to both sides, that eliminates -11x on the right
4x + 11x + 22 = 67
15x + 22 = 67
:
Subtract 22 from both sides, that eliminates 22 on the left
15x = 67 - 22
15x = 45
:
Divide both sides by 15, that leaves just x on the left
x = 45%2F15
x = 3
I leave it up to you to confirm these solutions, in this one substitute 3 for x in the original equation
:
:
43:
3x + 21 = -2x + 11
Follow the same rules
3x + 2x = 11 - 21
5x = -10
x = %28-10%29%2F5
x = -2
Check this solution the same way
:
45:
6x + 4 = 2x
6x - 2x = -4
4x = -4
x = %28-4%29%2F4
x = -1
:
Are you getting the idea now?