Verify $(y-x)y' = y-x+8$ has an explicit solution $y = x+ 4 \sqrt{x+2}$

Question

Ive solved this problem and both sides of the function don't appear to be equal to me. However the solution in the back of the book says it is a solution. Am I not simplifying enough? Thanks for the help!

(y-x)y' = y-x+8 ; y = x+ 4 sqrt(x+2)

                       y'=( 2/sqrt(x+2)) +1 

Then I plug in the values of y and y' into the function on the left.

(x+4*sqrt(x+2) -x)*(2/sqrt(x+2)) +1 = x+4*sqrt(x+2) -x +8

Answer 1

Step 1:

$$y = x+ 4 \sqrt{x+2} \implies y' = \dfrac{2}{\sqrt{x+2}}+1$$

Step 2:

$$(y-x)y' = (x+ 4 \sqrt{x+2} - x) \left(\frac{2}{\sqrt{x+2}}+1\right) = 4 \left(\sqrt{x+2}+2\right)$$

Step 3:

$$ y-x+8 = x+ 4 \sqrt{x+2} - x + 8 = 4 \left(\sqrt{x+2}+2\right)$$

Answer 2

We have $y'=\frac{y-x+8}{y-x}$ and a possible solution $y=x+4\sqrt{x+2}$ which gives $y'=\frac{2}{\sqrt{x+2}}+1$. Plugging this information into the DFQ we get: $$\frac{2}{\sqrt{x+2}}+1=\frac{x+4\sqrt{x+2}-x+8}{x+4\sqrt{x+2}-x}$$ which simplifies to: $$\frac{2}{\sqrt{x+2}}+\frac{\sqrt{x+2}}{\sqrt{x+2}}=\frac{4\sqrt{x+2}+8}{4\sqrt{x+2}}$$ When you divide right hand side term by factor 4, you get an identity. Conclusion?