Verified answer

Further to the other answers I think the answer to this problem is actually 1.

Here is my line of reasoning:

R = BCD + BC = BC(D + 1)

Now in Boolean Algebra, Anything + 1 = 1

Therefore

R = BC.1 = BC

Substitute this into the second equation

SBC = CD + BC

Divide both sides by BC

S = (CD + BC) / BC

S = CD/BC + BC/BC

S = CD/BC + 1

Again using the fact that Anything + 1 =1

S = 1

attempt posting this in Math particularly as I discovered it in math classification and it is not something programmers would study truly.. sure, that they had study it via commerce yet no longer in all probability via that call or something. do you decide on like (2>3)&&(a million==a million) = fake? or do you decide on like a=a million, b=0, a*b = 0, a+b = a million? boolean algebra is an entire branch of math and there are a number of categories, basically like there are distinctive styles of multiplication, so i choose anther occasion... your (x+y)(x+z) would not artwork except those variables have a value, in any different case it relatively is unsolvable, except you're doing certainty tables? Sorry, choose extra of a depiction. i'm going to examine back in the morning.

Yes divide SR by R then simplify.

R = BCD + BC

SR = CD + BC

Now interpret "R" value in the equation.

So..

S(BCD+BC) = CD+BC

Now bring (BCD+BC) to the other side

So finally......

S = (CD+BC) / (BCD+BC)