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 $C = \frac13$

Step-by-step explanation:

Given:

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

Rewriting $0=A+B$:

$0 = A +B \\ -A = B$

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

$1 = A -3B +C \\ 1 = A -3(-A) +C \\ 1 = A +3A +C \\ 1 = 4A +C \\ 1 -C = 4A \\ \frac{1 -C}{4} = A$

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

$-1 = -2A -2C \\ 2C -1 = -2A \\ \frac{2C -1}{-2} = A \\ -\frac{2C -1}{2} = A$

Since $A = -\frac{2C -1}{2}$ and $A = \frac{1 -C}{4}$ also, therefore $\frac{1 -C}{4} = -\frac{2C -1}{2}$. We can now solve for $C$ from the resulting equation.

Solving for $C$:

$\frac{1 -C}{4} = -\frac{2C -1}{2} \\ 1 -C = 2 \cdot -(2C -1) \\ 1 -C = -2(2C -1) \\ 1 -C = -4C +2 \\ -C = -4C +1 \\ 3C = 1 \\ C = \frac{1}{3}$

Solving for $A$:

$\frac{1 -C}{4} = A \\ \frac{1 -\frac{1}{3}}{4} = A \\ \frac{\frac{3}{3} -\frac{1}{3}}{4} = A \\ \frac{\frac{2}{3}}{4} = A \\ \frac{2}{3} \cdot \frac{1}{4} = A \\ \frac{2}{12} = A \\ \frac16 = A$

Solving for $B$:

$-A = B \\ -\frac16 = B$