Solve {l}{A+B=0}{C=4B}{A-4C=1}

$$\left\{ \begin{array} { l } { A + B = 0 } \\ { C = 4 B } \\ { A - 4 C = 1 } \end{array} \right.$$

Solve for A, B, C

$$A=\frac{1}{17}\approx 0.058823529$$

$$B=-\frac{1}{17}\approx -0.058823529$$

$$C=-\frac{4}{17}\approx -0.235294118$$

Short Steps Using Substitution

Reorder the equations.

$$C=4B A+B=0 A-4C=1$$

Substitute 4B for C in the equation A-4C=1.

$$A-4\times 4B=1$$

Solve the second equation for B and the third equation for A.

$$B=-A A=16B+1$$

Substitute -A for B in the equation A=16B+1.

$$A=16\left(-1\right)A+1$$

Solve A=16\left(-1\right)A+1 for A.

$$A=\frac{1}{17}$$

Substitute \frac{1}{17} for A in the equation B=-A.

$$B=-\frac{1}{17}$$

Substitute -\frac{1}{17} for B in the equation C=4B.

$$C=4\left(-\frac{1}{17}\right)$$

Calculate C from C=4\left(-\frac{1}{17}\right).

$$C=-\frac{4}{17}$$

The system is now solved.

$$A=\frac{1}{17} B=-\frac{1}{17} C=-\frac{4}{17}$$