Solve {l}{a+b+c=0}{-a-3b+c=0}{-2a-2c=-1}

$$\left. \begin{array} { l } { a + b + c = 0 } \\ { - a - 3 b + c = 0 } \\ { - 2 a - 2 c = - 1 } \end{array} \right.$$

Solve for a, b, c

$$a=1$$

$$b=-\frac{1}{2}=-0.5$$

$$c=-\frac{1}{2}=-0.5$$

Short Steps Using Substitution

Solve a+b+c=0 for a.

$$a=-b-c$$

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

$$-\left(-b-c\right)-3b+c=0 -2\left(-b-c\right)-2c=-1$$

Solve these equations for c and b respectively.

$$c=b b=-\frac{1}{2}$$

Substitute -\frac{1}{2} for b in the equation c=b.

$$c=-\frac{1}{2}$$

Substitute -\frac{1}{2} for c and -\frac{1}{2} for b in the equation a=-b-c.

$$a=-\left(-\frac{1}{2}\right)-\left(-\frac{1}{2}\right)$$

Calculate a from a=-\left(-\frac{1}{2}\right)-\left(-\frac{1}{2}\right).

$$a=1$$

The system is now solved.

$$a=1 b=-\frac{1}{2} c=-\frac{1}{2}$$