That a and b are additive inverses of one another.
A*(B-B) = A*0 = 0 Expanding the left hand side, using the distributive property, A*B + A*(-B) = 0 That is, A*B and A*(-B) are additive inverses. Next, (A-A)*(-B) = 0*(-B) = 0 Expanding, A*(-B) + (-A)*(-B) = 0 Therefore A*(-B) and (-A)*(-B) are additive inverses But, from above, the additive inverse of -A*B is A*B Therefore (-A)*(-B) = A*B It is not known when this was proven.
Real numbers are commutative under addition (and subtraction) so a + b - a = a - a + b The set of Real numbers includes an additive identity, 0, such that a - a = 0 so a - a + b = 0 + b The additive identity also has the property that 0 + b = b [= b + 0] so 0 + b = b
the answer is a
4b
Yes because if 1+0=1 than 0 plus b equals b
a=24 b=16 c=18
Because there is no way to define the divisors, the equations cannot be evaluated.
a= (+a) or a= (-) b= 2a b= 2a c= (-a) c= (+a)
A.
a+b=a+b
a = 20 b = 60 c = 100