Solve {l}{3a+3b+c=0}{-4a+b-c=0}{3a+b+3c=-1}

$$\left. \begin{array} { l } { 3 a + 3 b + c = 0 } \\ { - 4 a + b - c = 0 } \\ { 3 a + b + 3 c = - 1 } \end{array} \right.$$

Solve for a, b, c

$$a=\frac{1}{8}=0.125$$

$$b=\frac{1}{32}=0.03125$$

$$c=-\frac{15}{32}=-0.46875$$

Short Steps Using Substitution

Solve 3a+3b+c=0 for c.

$$c=-3a-3b$$

Substitute -3a-3b for c in the second and third equation.

$$-4a+b-\left(-3a-3b\right)=0 3a+b+3\left(-3a-3b\right)=-1$$

Solve these equations for b and a respectively.

$$b=\frac{1}{4}a a=-\frac{4}{3}b+\frac{1}{6}$$

Substitute \frac{1}{4}a for b in the equation a=-\frac{4}{3}b+\frac{1}{6}.

$$a=-\frac{4}{3}\times \frac{1}{4}a+\frac{1}{6}$$

Solve a=-\frac{4}{3}\times \frac{1}{4}a+\frac{1}{6} for a.

$$a=\frac{1}{8}$$

Substitute \frac{1}{8} for a in the equation b=\frac{1}{4}a.

$$b=\frac{1}{4}\times \frac{1}{8}$$

Calculate b from b=\frac{1}{4}\times \frac{1}{8}.

$$b=\frac{1}{32}$$

Substitute \frac{1}{32} for b and \frac{1}{8} for a in the equation c=-3a-3b.

$$c=-3\times \frac{1}{8}-3\times \frac{1}{32}$$

Calculate c from c=-3\times \frac{1}{8}-3\times \frac{1}{32}.

$$c=-\frac{15}{32}$$

The system is now solved.

$$a=\frac{1}{8} b=\frac{1}{32} c=-\frac{15}{32}$$