Solve {l}{0=4x+2B+C}{-3=-1x-1B+C}{0=4x-2B+C}

$$\left\{ \begin{array} { l } { 0 = 4 x + 2 B + C } \\ { - 3 = - 1 x - 1 B + C } \\ { 0 = 4 x - 2 B + C } \end{array} \right.$$

Solve for x, B, C

$$x=\frac{3}{5}=0.6$$

$$B=0$$

$$C = -\frac{12}{5} = -2\frac{2}{5} = -2.4$$

Short Steps Using Substitution

Solve 0=4x+2B+C for C.

$$C=-4x-2B$$

Substitute -4x-2B for C in the second and third equation.

$$-3=-x-B-4x-2B 0=4x-2B-4x-2B$$

Solve these equations for x and B respectively.

$$x=\frac{3}{5}-\frac{3}{5}B B=0$$

Substitute 0 for B in the equation x=\frac{3}{5}-\frac{3}{5}B.

$$x=\frac{3}{5}-\frac{3}{5}\times 0$$

Calculate x from x=\frac{3}{5}-\frac{3}{5}\times 0.

$$x=\frac{3}{5}$$

Substitute \frac{3}{5} for x and 0 for B in the equation C=-4x-2B.

$$C=-4\times \frac{3}{5}-2\times 0$$

Calculate C from C=-4\times \frac{3}{5}-2\times 0.

$$C=-\frac{12}{5}$$

The system is now solved.

$$x=\frac{3}{5} B=0 C=-\frac{12}{5}$$