Solve {l}{0=A+B}{0=8A+2B+C}{1=16A-8B-2C}

$$\left. \begin{array} { l } { 0 = A + B } \\ { 0 = 8 A + 2 B + C } \\ { 1 = 16 A - 8 B - 2 C } \end{array} \right.$$

Solve for A, B, C

$$A=\frac{1}{36}\approx 0.027777778$$

$$B=-\frac{1}{36}\approx -0.027777778$$

$$C=-\frac{1}{6}\approx -0.166666667$$

Short Steps Using Substitution

Solve 0=A+B for A.

$$A=-B$$

Substitute -B for A in the second and third equation.

$$0=8\left(-1\right)B+2B+C 1=16\left(-1\right)B-8B-2C$$

Solve these equations for B and C respectively.

$$B=\frac{1}{6}C C=-12B-\frac{1}{2}$$

Substitute \frac{1}{6}C for B in the equation C=-12B-\frac{1}{2}.

$$C=-12\times \frac{1}{6}C-\frac{1}{2}$$

Solve C=-12\times \frac{1}{6}C-\frac{1}{2} for C.

$$C=-\frac{1}{6}$$

Substitute -\frac{1}{6} for C in the equation B=\frac{1}{6}C.

$$B=\frac{1}{6}\left(-\frac{1}{6}\right)$$

Calculate B from B=\frac{1}{6}\left(-\frac{1}{6}\right).

$$B=-\frac{1}{36}$$

Substitute -\frac{1}{36} for B in the equation A=-B.

$$A=-\left(-\frac{1}{36}\right)$$

Calculate A from A=-\left(-\frac{1}{36}\right).

$$A=\frac{1}{36}$$

The system is now solved.

$$A=\frac{1}{36} B=-\frac{1}{36} C=-\frac{1}{6}$$