0=A+B 1=A-3B+C -1=-2A-2C Find the values of A, B and C

Solving Systems of 3 EquationsAnswer:[tex]A = \frac16\\[/tex], [tex]B = -\frac16\\[/tex] and [tex]C = \frac13\\[/tex]Step-by-step explanation:Given:[tex]&0 = A +B \\ &1 = A -3B +C \\ &-1 = 2A -2C[/tex]Rewriting [tex]0=A+B[/tex]:[tex]0 = A +B \\ -A = B[/tex]Rewriting [tex]1=A-3B+C[/tex]:[tex]1 = A -3B +C \\ 1 = A -3(-A) +C \\ 1 = A +3A +C \\ 1 = 4A +C \\ 1 -C = 4A \\ \frac{1 -C}{4} = A[/tex]Rewriting [tex]-1=-2A-2C[/tex]:[tex]-1 = -2A -2C \\ 2C -1 = -2A \\ \frac{2C -1}{-2} = A \\ -\frac{2C -1}{2} = A[/tex]Since [tex]A = -\frac{2C -1}{2}\\[/tex] and [tex]A = \frac{1 -C}{4}\\[/tex] also, therefore [tex]\frac{1 -C}{4} = -\frac{2C -1}{2}\\[/tex]. We can now solve for [tex]C[/tex] from the resulting equation.Solving for [tex]C[/tex]:[tex]\frac{1 -C}{4} = -\frac{2C -1}{2} \\ 1 -C = 2 \cdot -(2C -1) \\ 1 -C = -2(2C -1) \\ 1 -C = -4C +2 \\ -C = -4C +1 \\ 3C = 1 \\ C = \frac{1}{3}[/tex]Solving for [tex]A[/tex]:[tex]\frac{1 -C}{4} = A \\ \frac{1 -\frac{1}{3}}{4} = A \\ \frac{\frac{3}{3} -\frac{1}{3}}{4} = A \\ \frac{\frac{2}{3}}{4} = A \\ \frac{2}{3} \cdot \frac{1}{4} = A \\ \frac{2}{12} = A \\ \frac16 = A[/tex]Solving for [tex]B[/tex]:[tex]-A = B \\ -\frac16 = B [/tex]