Given the system of equations:
2x-4y=10
x+5y=40
Using the substitution method:
x=40-5y (isolating one of the variables)
2(40-5y) -4y = 10
80 -10y-4y=10
-14y=10-80
-14y=-70
y=5
Back to the initial equation:
x=40-5y
x=40-5*5
x=40-25
x=15
So, the solution to the system is S={15;5}
