Question

How do you solve `x/3=2/3x+1/6`?

Answer 1

`x=-1/2`

Explanation:

The equation is `1/3x=2/3x+1/6`

Bringing x terms to one side:

`1/3x-2/3x=1/6`

`-1/3x=1/6`

`-x=3/6`

`x=-1/2`

Answer 2

1) `(1x)/3 = (2x)/3 + 1/6`
2) `hArr (1x)/3 - (2x)/3 - 1/6 = 0`
3) `hArr (2x)/6 - (4x)/6 - 1/6 = 0`
4) `hArr (2x - 4x -1)/6 = 0`
5) `hArr (-2x - 1)/6 = 0`
6) `rArr -2x - 1 = 0`
7) `hArr -2x = 1`
8) `hArr x = -1/2`

Explanation:

1) Your equation.
2) I put everything in the same side.
3) I put everything to the same denominator.
4) I put everything on the same fraction bar.
5) I reduced the common terms.
6) « If a fraction is equal to zero, then the numerator is equal to zero. » That's a propriety.
7) I isolated the unknown value.
8) Same.

Sorry for bad English... I hope this helped you. :)

Answer 3

My first impulse would be to multiply everything by 3, just to get rid of the fractional `x`'s.

Explanation:

`->x=2x+1/2`

Now subtract `2x` from both sides:

`->x-2x=cancel(2x)-cancel(2x)+1/2`

`->-x=1/2->x=-1/2`