by transitive property
If a=b and c=d then (a+c)=(b+d) ? This is proved very simply by the direct application of perhaps the most fundamental statement in all of Algebra: "If equals are added to equals, the sums are equal."
a = 20 b = 60 c = 100
a=24 b=16 c=18
2s-2b= a+b+c-2b simplified that would be a+c-b.
a - b = c -(a - b) = -c b - a = -c
This is true. As it is the same number for A and B so taking C from one would be the same as taking C from the other.
a= (+a) or a= (-) b= 2a b= 2a c= (-a) c= (+a)
2a. (a, b and c are all equal.)
A.
A=B , A-B=B-B , A-B =0 B=C , B-B=C-B, 0=C-B So A-B=0 but also C-B=0 A-B=C-B ...add +b ...A-B+B=C-B+B , A=C
by transitive property
Yes.
If a=b and c=d then (a+c)=(b+d) ? This is proved very simply by the direct application of perhaps the most fundamental statement in all of Algebra: "If equals are added to equals, the sums are equal."
a = 20 b = 60 c = 100
A=0 b=0 c=0
The transitive property of equality says that if a=b, then b=c.If a=b and b=c, then a=cTo Prove:Using the equation:a=bsubstituting the value of b in terms of c:which is: b=ctherefore:a=ba=(c)a=c