SOLUTION: I am having problems in solving this equation. 19-(2x+3)=2(x+3)+x take everthing out of bracket 19-2x+3=2x+6+x without Bracket +2x+22=2x+6+x Added 19+3 +22=-2x+2x+x move varia

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Question 16310: I am having problems in solving this equation.
19-(2x+3)=2(x+3)+x take everthing out of bracket
19-2x+3=2x+6+x without Bracket
+2x+22=2x+6+x Added 19+3
+22=-2x+2x+x move variable to one side
+22=x -2x+2x cancel each other out
What am I doing wrong
Thank-you
I'm sorry I gave you the wrong e-mail address here is the correct one.
Found 2 solutions by elima, pwac:
Answer by elima(1433) (Show Source):
You can put this solution on YOUR website!
19-(2x+3)=2(x+3)+x take everthing out of bracket
19-2x+3=2x+6+x without Bracket
+2x+22=2x+6+x Added 19+3
+22=-2x+2x+x move variable to one side
+22=x -2x+2x cancel each other out
First when you distribute the -1 to (2x+3), they both change to -;
19-2x-3=2(x+3)+x
Now distribute the 2(x+3);
which you did correctly;
so you have:
19-2x-3=2x+6+x
Now move variable to one side and add like terms;
16-6=2x+x+2x
10=5x
Divide by 5;
=
2=x
=)

Answer by pwac(253) (Show Source):
You can put this solution on YOUR website!
method is correct but you made simple signing error.
also 6 mysteriously disappeared between 3rd and 4th line.
19-(2x+3)=2(x+3)+x ok remove brackets
19-2x-3=2x+6+x note sign changes on 3.you are actually doing 19-(+3)
16-2x=3x+6 you have summed like terms here
now collect like terms on same side(don't forget the 6)
also when they change sides they change sign
16-6=3x+2x
10=5x divide both sides by 5
2=x
or
x=2 same thing
substitute x in original with 2 to check
hope this helps
Pete