Question

How do you solve `1/6-x/2=(x-5)/3`?

Answer 1

see the entire solution process below:

Explanation:

First, multiple each side of the equation by `color(red)(6)` (the lowest common denominator of all the fractions) to eliminate the fractions while keeping the equation balanced:

`color(red)(6)(1/6 - x/2) = color(red)(6) xx (x - 5)/3`

`(color(red)(6) xx 1/6) - (color(red)(6) xx x/2) = cancel(color(red)(6))2 xx (x - 5)/color(red)(cancel(color(black)(3)))`

`(cancel(color(red)(6)) xx 1/color(red)(cancel(color(black)(6)))) - (cancel(color(red)(6))3 xx x/color(red)(cancel(color(black)(2)))) = 2(x - 5)`

`1 - 3x = 2x - 10`

Next, add `color(red)(3x)` and `color(blue)(10)` to each side of the equation to isolate the `x` term while keeping the equation balanced:

`1 - 3x + color(red)(3x) + color(blue)(10) = 2x - 10 + color(red)(3x) + color(blue)(10)`

`1 + color(blue)(10) - 3x + color(red)(3x) = 2x + color(red)(3x) - 10 + color(blue)(10)`

`11 - 0 = 5x - 0`

`11 = 5x`

Now, divide each side of the equation by `color(red)(5)` to solve for `x` while keeping the equation balanced:

`11/color(red)(5) = (5x)/color(red)(5)`

`11/5 = (color(red)(cancel(color(black)(5)))x)/cancel(color(red)(5))`

`11/5 = x`

`x = 11/5`

Answer 2

`x=11/5`

Explanation:

Eliminate the fractions in the equation by multiplying ALL terms on both sides of the equation by the `color(blue)"lowest common multiple"` ( LCM ) of the denominators 6, 2 and 3

The LCM of 6 , 2 and 3 is 6

Hence multiply All terms by 6

`(cancel(6)^1xx1/cancel(6)^1)-(cancel(6)^3xx x/cancel(2)^1)=(cancel(6)^2xx(x-5)/cancel(3)^1)`

`rArr1-3x=2(x-5)larr" no fractions"`

distribute the bracket on the right side.

`rArr1-3x=2x-10`

subtract 2x from both sides.

`1-3x-2x=cancel(2x)cancel(-2x)-10`

`rArr1-5x=-10`

subtract 1 from both sides.

`cancel(1)cancel(-1)-5x=-10-1`

`rArr-5x=-11`

To solve for x, divide both sides by - 5

`(cancel(-5) x)/cancel(-5)=(-11)/(-5)`

`rArrx=11/5`

`color(blue)"As a check"`

substitute this value into the equation and if the left side equals the right side then it is the solution.

`"left side "=1/6-(11/5)/2=1/6-11/10=5/30-33/30=-28/30=-14/15`

`"right side"=(11/5-5)/3=(11/5-25/5)/3=(-14/5)/3=-14/15`

`rArrx=11/5" is the solution"`