Question

How do you solve `5/9x+3=2/3`?

Answer 1

See the entire solution process below:

Explanation:

First, multiply each side of the equation by `color(red)(9)` to eliminate the fraction while keeping the equation balanced. Eliminating the fraction will make the problem easier to work with.

`color(red)(9)(5/9x + 3) = color(red)(9) xx 2/3`

`(color(red)(9) xx 5/9x) + (color(red)(9) xx 3) = cancel(color(red)(9))3 xx 2/color(red)(cancel(color(black)(3)))`

`(cancel(color(red)(9)) xx 5/color(red)(cancel(color(black)(9)))x) + 27 = 6`

`5x + 27 = 6`

Next, subtract `color(red)(27)` from each side of the equation to isolate the `x` term while keeping the equation balanced:

`5x + 27 - color(red)(27) = 6 - color(red)(27)`

`5x + 0 = -21`

`5x = -21`

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

`(5x)/color(red)(5) = -21/color(red)(5)`

`(color(red)(cancel(color(black)(5)))x)/cancel(color(red)(5)) = -21/5`

`x = -21/5`

Answer 2

`x=-21/5`

Explanation:

A fraction consists of: `" "("count")/("size indicator") -> ("numerator")/("denominator")`

You can not directly subtract or add the counts (numerators) unless the size indicators (denominators) are the same.

`color(green)((5x)/9+[3color(red)(xx1)]=2/3color(red)(xx1)`

`color(green)((5x)/9+[3color(red)(xx9/9)]=2/3color(red)(xx3/3)`

`color(green)((5x)/9+[27/9]=6/9)`

As all the denominators (size indicators) are the same it is equally true that:

`5x+27=6`

Subtract 27 from both sides

`5x=-21`

Divide both sides by 5

`x=-21/5`