First of all we need to find the values of x and y satisfying these inequalities.
Solving inequality for x
$$x^{2} - 5x + 4 <= 0$$
$$x^2\ -\ 4x\ -\ x\ +\ 4\le0$$
$$x\left(x\ -\ 4\right)\ -1\left(\ x\ -\ 4\right)\le0$$
$$\left(x-1\right)\left(\ x\ -\ 4\right)\le0$$
equation has 2 roots ie. x=1 and x=4
$$x^{2} - 5x + 4 <= 0$$ has solution [1,4]
and now solving the second inequality for the values of y which satisfy $$y^{2} -6y+5 <= 0$$
$$y^2-5y-y+5\le0$$
$$y\left(y-5\right)-1\left(y-5\right)\le0$$
$$\left(y-1\right)\left(y-5\right)\le0$$
As this equation has roots at y=1 and 5 the satisfying values of this inequality will be [1,5].
Moving on to the question we have to find maximum and minimum value of$$\frac{y+7x}{y+4x}$$
which can be written as 1 + $$\frac{3x}{y+4x}$$,
which also equals 1 + $$\frac{3}{y/x+4}$$.
This reaches its maximum when $$\frac{y}{x}$$ is at its minimum and vice versa.
Maximum value of y/x = 5/1
Minimum value of y/x = 1/4
So, the minimum value of $$\frac{y+7x}{y+4x}$$ is $$1+\frac{3}{5+4}$$ =$$\frac{4}{3}$$ = 12/9
and the maximum is$$1+\frac{3}{\frac{1}{4}+4}$$ = $$1+\frac{12}{17}$$=$$\frac{29}{17}$$
