linear equations: solving x/2 + 3 = 6

crappiefisher26

New member
Joined
Jul 23, 2006
Messages
49
solving linear equation with several occurences of the variable

Here is the question:

. . .(x/2) + 3 = 6

. . .2(x/2) + 2(3/1) = 2(6/1)

. . .x + 2 = 6

Subtract 2 from both sides:

. . .x = 4

Then I plugged "4" in for "x" in the original equation, but it didn't work. Could someone let me know what I did wrong? Thank you.
 

galactus

Super Moderator
Staff member
Joined
Sep 28, 2005
Messages
7,216
Re: linear equations

crappiefisher26 said:
solving linear equation with several occurences of the variable

Ok heres the question

(x/2)+3=6

2(x/2)+2(3/1)=Here2(6/1)

x+6=12

x+2=6

subtract 2 from both sides

x=4, i plugges 4 in for x in the original equation but it didnt work, so could someone let me know what i did wrong.
 

crappiefisher26

New member
Joined
Jul 23, 2006
Messages
49
linear equation

I must be stupid or somethin.

(x/3)-2=1

6(x/3)-6(2/1)=6(1/1)

2x-3=6

add -3 on each side

2x=3

would the answer be x=(3/2) that doesnt work if ya plug it back in though
 

galactus

Super Moderator
Staff member
Joined
Sep 28, 2005
Messages
7,216
You can look at (x/3)-2=1 and tell the answer is x=9, without algebra.

But to go through the motions:

3(x/3)-3(2)=3(1)

x-6=3

x=9
 

stapel

Super Moderator
Staff member
Joined
Feb 4, 2004
Messages
15,943
crappiefisher26 said:
6(x/3) - 6(2/1) = 6(1/1)

2x - 3 = 6
Six times two is not three.

Eliz.
 

crappiefisher26

New member
Joined
Jul 23, 2006
Messages
49
yeh i saw that after i posted. Sorry
 
Top