Abc

$3A=30$ $\implies$ $A=\dfrac{30}{3}$ $\implies$ $A=10$

$A+2B=18$

Substituting $10$ to $A$ in the above equation, we get

$10+2B=18$ $\implies$ $2B=18-10$ $\implies$ $2B=8$ $\implies$ $B=4$

$B-C=2$

Substituting $4$ to $B$ in the above equation, we get

$4-C=2$ $\implies$ $C=2$

Finally,

$A+B+C=10+4+2=$ $\boxed{\color{#D61F06}16}$

Given that,

A + A + A = 30

A + B+ B = 18

B - C = 2

Now,

A + A + A = 30

$⇒$ 3A = 30

∴ A = 10 ;[Dividing 30 with 3]

Now place the value of A to get B

A + B+ B = 18

$⇒$ 10 + 2B = 18

$⇒$ 2B = 18 - 10

∴ B = 4 ;[ Dividing 8 with 2]

Now place the value of B to get C

B - C = 2

$⇒$ 4 - C = 2

∴ C = 2 ; [subtracting 2 with 4]

Therefore, A + B + C $⇒$ 10 + 4 + 2 = 16

A+A+A=3A

so,3A=30

or,A= $\frac{30}{3}$

                or,A=10

AGAIN

or,10+B+B=18

or,10+2B=18

or,2B=18-10

or,2B=8

                 or,B=4

NOW,

or,B - C = 2

or,4-C=2

or,-C=2-4

or,-C=-2

               or,C=2

so,A+B+C

=10+4+2

=16(ANSWER)

3A=30, or, A=10.................................(1)

again, 10+2B=18, or,B=4....................(2)

and,4-C=2, or,C=2...............................(3)

so,A+B+C=10+4+2

                  =16

Because B - C = 2, we know that C - B = -2. We can add this to the second equation, A + B + B = 18, to get A + B + C = 16.

(A+B+B)-(B-C)=18-2
A+B+B-B+C=A+B+C
=16