Given the system:

a = 2+ b .............(1)

b = 3c .............(2)

a+b = c ...............(3)

We will use the substitution method to solve.

We will substitute (2) into (1).

==> a= 2+ b = 2 + 3c

==> a = 2+ 3C ..............(4)

Now we will substitute (2) and (3) into (4).

==> a + b = C

==> (2+3c) + 3c = c

==> 2 + 6c = c

==> 5c = -2

==> c = -2/5

==> b = 3c = 3*-2/5 = -6/5

==> b = -6/5

==> a = 2+ b = 2 - 6/5 = 4/5

==> a = 4/5

**Then, the answer is:**

**a= 4/5 b= -6/5 c = -2/5**

a = 2+b...(1).

b= 3c.......(2).

a+b = c....(3). What are a, band c.

We put a= 2+b in a+b=c.

2+b+b = c.

=> c= 2+2b. So we purt this value of c in (2):

b= 3(2+2b).

=> b = 6+6b, so 5b = -6,** b= -6/5.**

Put b = -6/6 in (1): a= 2+b = 2+(-6/5) = 4/5. a = 4/5.

Put a = 4/5 and b= -6/5 in (3): a+b= c, Or c= a+b = (4/5-6/5 = -2/5.

Therefore a= 4/5, b= -6/5, and c = -2/5.

a = 2+b

a - b = 2 (1)

b= 3c => c = b/3 (2)

a+b = c => a+b = b/3

We'll subtract b/3

a + b - b/3 = 0

a + 2b/3 = 0

3a + 2b = 0 (3)

We'll add 2(1) + (3):

2a - 2b + 3a + 2b = 4

We'll eliminate like terms:

5a = 4

**a = 4/5**

4/5 = 2 + b

b = 4/5 - 2

**b = -6/5**

**c = -2/5**