If the graphs of 7x+5y=35 and 4x+8y=32 are plotted, the lines cross at (10/3,7/3). The area between the lines after the point of intersection is the triangular area that satisfies the two linear inequalities bound by the x axis, to satisfy the condition that y>0. The minimum value of z is related to the intersection point of the two lines and the triangular bound area. At the point of intersection z=27*10/3+13*7/3=90+91/3=120 1/3 or 361/3 or 120.33.
The point of intersection of the lines is found from the equations 7x+5y=35 and 4x+8y=32 (x+2y=8). So x=8-2y and this can be substituted into the other equation: 56-14y+5y=35; 9y=21 so y=7/3 and x=8-14/3=10/3. The areas on the right (positive) side of the linear equalities represent the solution to each equation, so the common triangular area on the right represents the solutions for (x,y) where both inequality equations are satisfied and where y>0 is satisfied.
The triangular area is bounded on its left side by the line 7x+5y=35, so y=7-1.4x. Substitute this value in the z equation: z=27x+13(7-1.4x)=8.8x+91. The minimum value of z is when x is at a minimum, which is at the point of intersection where x=10/3, so z=361/3=120.33.
Note that, if I had used a substitution for x=5-5y/7, so that z=27(5-5y/7)+13y, at the minimum value of y=0, z=135, which is greater than 120.33. The coefficient of x in the z equation has greater significance than the coefficient of y. If z=24x+13y, then when y=0, z=24(5-5y/7)+13y=120, and this would have been the minimum z.