# Solving linear equation.

• Apr 6th 2010, 06:30 PM
hoiguy1208
Solving linear equation.
3(g-3)=6

How do I do these problems?

(BTW: I understood the things of the 4 variable things).
• Apr 6th 2010, 06:34 PM
Stroodle
To solve for $g$

You can first divide both sides of the equation by 3, giving:

$g-3=2$

then add 3 to both sides:

$g=5$
• Apr 6th 2010, 06:53 PM
hoiguy1208
Attempt at another problem
3(x+1)=21
-3 -3

x+1=18
-1 -1

x=17?

The numbers are aligned kind of wrong, but you get the idea.
• Apr 6th 2010, 07:47 PM
xSquall
$3(g-3)=6 \mbox{ / in both sides } \frac{1}{3}$
$\Leftrightarrow{\frac{3(g-3)}{3}=\frac{6}{3}}$
$\Leftrightarrow{(g-3)=2}\mbox{ / in both sides plus +3}$
$\Leftrightarrow{g-3+3=2+3}$
$\Leftrightarrow{g=5} \mbox{ end}$
• Apr 6th 2010, 10:17 PM
BabyMilo
Quote:

Originally Posted by hoiguy1208
3(x+1)=21
-3 -3

x+1=18
-1 -1

x=17?

The numbers are aligned kind of wrong, but you get the idea.

Step 1
3(x+1)=3x+3 (multiply out the bracket)

Step 2
3x+3=21 (-3 from both side)
3x=18

Step 3
3x=18 (divide both side by 3)
x=6.

does this help?
• Apr 7th 2010, 12:32 AM
hoiguy1208
Thank you. On problems like:

6(3d+5) =75, do I take a similar approach?
• Apr 7th 2010, 12:39 AM