Let burger be represented with b, and order of fries be represented with f.
So, the system of equations are;
5b + 3f = 35.50 .....eqn(1)
2b + 6f = 25.00 ......eqn(2)
By elimination method of solving simultaneous equations,
2(5b + 3f = 35.50) ; 10b + 6f = 71.00 .....eqn(3)
5(2b + 6f = 25.00) ; 10b + 30f = 125.00 .....eqn(4)
Subtracting eqn(3) from eqn(4), we get,
24f = 54
Dividing both sides by 24, we get,
f = 54/24 = 2.25 =$2.25
To find the value of b:
We substitute 2.25 for f in eqn(2):
2b + 6(2.25) = 25
2b = 25 - 13.50
2b = 11.50
Dividing both sides by 2, we get,
b =11.50/2
b = 5.75 = $5.75
The solution is;
the cost of a burger is $5.75