Question

How do you solve `(x+2)/3+(x-3)/4=1`?

Answer 1

See the entire solution process below:

Explanation:

First, we need to eliminate the fractions by multiplying each side of the equation by a common denominator for the two fractions.

`3 x 4 = color(red)(12)` is a common denominator.

`color(red)(12)((x + 2)/3 + (x - 3)/4) = color(red)(12) xx 1`

`(color(red)(12) xx (x + 2)/3) + (color(red)(12) xx (x - 3)/4) = 12`

`(cancel(color(red)(12)) 4 xx ((x + 2))/color(red)(cancel(color(black)(3)))) + (cancel(color(red)(12)) 3 xx ((x - 3))/color(red)(cancel(color(black)(4)))) = 12`

`4(x + 2) + 3(x - 3) = 12`

We can now expand the terms in parenthesis and group and combine like terms on the left side of the equation:

`(4 xx x) + (4 xx 2) + (3 xx x) - (3 xx 3) = 12`

`4x + 8 + 3x - 9 = 12`

`4x + 3x + 8 - 9 = 12`

`7x - 1 = 12`

Next, we can add `color(red)(1)` to each side of the equation to isolate the `x` term while keeping the equation balanced:

`7x - 1 + color(red)(1) = 12 + color(red)(1)`

`7x - 0 = 13`

`7x = 13`

Now, divide each side of the equation by `color(red)(7)` to solve for `x` while keeping the equation balanced:

`(7x)/color(red)(7) = 13/color(red)(7)`

`(color(red)(cancel(color(black)(7)))x)/cancel(color(red)(7)) = 13/7`

`x = 13/7`