The transitive property of equality says that if a=b, then b=c.
If a=b and b=c, then a=c
To Prove:
Using the equation:
a=b
substituting the value of b in terms of c:
which is: b=c
therefore:
a=b
a=(c)
a=c
a - b = c -(a - b) = -c b - a = -c
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
Yes.
To show this proof, you must use substitution. If A=B and B=C, you can substitute C wherever you see B. By doing this, you can write the first equation as A=C.
Pascal
a - b = c -(a - b) = -c b - a = -c
a= (+a) or a= (-) b= 2a b= 2a c= (-a) c= (+a)
2a. (a, b and c are all equal.)
A.
by transitive property
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
Yes.
a = 20 b = 60 c = 100
A=0 b=0 c=0
a=24 b=16 c=18