[tex]3x+\frac{1}{10}=x-\frac{1}{7}[/tex]
To solve for 'x' we first collect simmilar terms. In this case I'll collect 'x'-terms on the left and non-'x'-terms on the right.
Start by subtracting 1/10 from both sides:
[tex]3x+\frac{1}{10}-\frac{1}{10}=x-\frac{1}{7}-\frac{1}{10}[/tex]
and simplify:
[tex]3x=x-\frac{10}{70}-\frac{7}{70}[/tex]
[tex]3x=x-\frac{17}{70}[/tex]
We now have all non-'x'-terms on the right. We can get all the 'x' terms on the left by subtracting 'x' from both sides:
[tex]3x-x=x-\frac{17}{70}-x[/tex]
and simplifying:
[tex]2x=-\frac{17}{70}[/tex]
Then we can solve for x by dividing both sides by 2:
[tex]\frac{2x}{2}=\frac{-\frac{17}{70}}{2}[/tex]
and simplifying:
[tex]x=-\frac{17}{140}[/tex]
