Question

How do you solve for u in `y= (u+1)/(u+2)`?

Answer 1

`color(blue)(u = (1 -2y)/(y-1)`

Explanation:

`y= (u+1) / color(green)((u+2)`

`y *color(green)( (u+2)) = (u+1)`

`y * color(green)( (u)) + y * color(green)((2)) = (u+1)`

`yu + 2y = u+1`

Isolating `u` on the L.H.S

`ycolor(blue)(u) -color(blue)( u) = 1 -2y`

`u` is common to both terms of the L.H.S:

`color(blue)(u) (y-1) = 1 -2y`

`color(blue)(u = (1 -2y)/(y-1)`

Answer 2

`u=(-2y+1)/(y-1)`

Explanation:

Another (more difficult) method:

Rewrite `u+1` as `u+2-1`.

`y=(u+2-1)/(u+2)`

Split up the numerator.

`y=(u+2)/(u+2)-1/(u+2)`

`y=1-1/(u+2)`

Rearrange the terms.

`y-1=-1/(u+2)`

Multiply both sides by `(u+2)`.

`(y-1)(u+2)=-1`

Distribute the `(y-1)` into `u` and `2`:

`u(y-1)+2(y-1)=-1`

`u(y-1)+2y-2=-1`

`u(y-1)=-2y+1`

After this algebraic rearrangement, divide both sides by `(y-1)` to isolate `u`.

`u=(-2y+1)/(y-1)`