Question

0=A+B
1=A-3B+C
-1=-2A-2C

Find the values of A, B and C

Answer 1

Solving Systems of 3 Equations

Answer:

`A = \frac16\\`,
`B = -\frac16\\` and
`C = \frac13\\`

Explanation:

Given:

`&0 = A +B \\ &1 = A -3B +C \\ &-1 = 2A -2C`

Rewriting
`0=A+B`:

`0 = A +B \\ -A = B`

Rewriting
`1=A-3B+C`:

`1 = A -3B +C \\ 1 = A -3(-A) +C \\ 1 = A +3A +C \\ 1 = 4A +C \\ 1 -C = 4A \\ (1 -C)/(4) = A`

Rewriting
`-1=-2A-2C`:

`-1 = -2A -2C \\ 2C -1 = -2A \\ (2C -1)/(-2) = A \\ -(2C -1)/(2) = A`

Since
`A = -(2C -1)/(2)\\` and
`A = (1 -C)/(4)\\` also, therefore
`(1 -C)/(4) = -(2C -1)/(2)\\`. We can now solve for
`C` from the resulting equation.

Solving for
`C`:

`(1 -C)/(4) = -(2C -1)/(2) \\ 1 -C = 2 \cdot -(2C -1) \\ 1 -C = -2(2C -1) \\ 1 -C = -4C +2 \\ -C = -4C +1 \\ 3C = 1 \\ C = (1)/(3)`

Solving for
`A`:

`(1 -C)/(4) = A \\ (1 -(1)/(3))/(4) = A \\ ((3)/(3) -(1)/(3))/(4) = A \\ ((2)/(3))/(4) = A \\ (2)/(3) \cdot (1)/(4) = A \\ (2)/(12) = A \\ \frac16 = A`

Solving for
`B`:

`-A = B \\ -\frac16 = B`