Verified answer

what you're presenting is a system of 2 equations with two unknowns. This usually has one solution for (v,b).

Constraining v and b further saying that it they must have the same value is throwing in another equation.

Then you have 2 equations with two unknowns. If these 3 equations are not linearly dependents then there will not be a solution.

There is no other recourse then to check. You can do this with matrix algebra or just with the old tricks you have learned:

10v+4b=272

1v+11b=377

doing 10 times the second equation and subtracting it from the first equation and replace this result for the second equation gives us the following:

10v+4b=272 (let's leave the first equation)

0v-106b=-3498

so b = 33

now let's substitute b in the first equation:

10v + 132 =272

v = (272-132)/10 =14

so. b and v are different. This system of equations cannot be satisfied if b and v have to be the same amount.

v = 14, b = 33