A graphical solution is probably the easiest way to solve this LP problem.
If we plot the graphs of 5x+4y=20 and 2x+3y=12 we inspect the quadrilateral making up the feasible region. By shading the area between each line and the x and y axes, we can identify the region where the shadings overlap, the feasible region satisfying the constraints.
The lines intersect at the point which is a solution of the system of two equations:
5x+4y=20 multiply by 3: 15x+12y=60,
2x+3y=12 multiply by 4: 8x+12y=48.
Subtract: 7x=12, x=12/7, 3y=12-2x=12-24/7=60/7, y=20/7.
Intersection at (12/7,20/7).
4 vertices are shown below with the objective function P=5x+7y calculated for each vertex:
(0,0) P=0
(0,4) P=28
(4,0) P=20
(12/7,20/7) P=60/7+140/7=200/7=28.6.
P is max when x=12/7 and y=20/7.
Double-check the constraints: 5x+4y=60/7+80/7=20; 2x+3y=24/7+60/7=12.
Unless there's an error in the question (< should be ≤), the constraints are not validated, but only just! Strictly P is maximum when x<12/7 and y<20/7 by the smallest possible amount.