What is the maximum value of $\frac{7x+2y}{2x+2}+\frac{3x+8y}{2y+2}$ while $0≤x,y≤1$

I found a solution myself with helps of the comments above on the question.
The solution is similar with this solution of my previous question.
Since $x,y\leq1$ $$\dfrac{7x+2y}{2x+2}\leq\dfrac{7x+2y}{2x+2y}\\ \dfrac{3x+8y}{2y+2}\leq\dfrac{3x+8y}{2x+2y}$$ so $$ \dfrac{7x+2y}{2x+2}+\dfrac{3x+8y}{2y+2}\leq\dfrac{10x+10y}{2x+2y}=5$$

Equality will hold when $x=1$ & $y=1$

$$ f(x,y) = \frac{7x+2y}{2x+2} + \frac{3x+8y}{2y+2} = \left( \frac 72 - \frac{7-2y}{2x+2} \right)+ \left( 4 - \frac{8 - 3x}{2y+2} \right) $$ is strictly increasing in both variables on $[0, 1]^2$, therefore $$ f(x, y) \le f(1,1) = 5 $$ with equality exactly for $(x, y) = (1,1)$.