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

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

Solve for a, b, c

$$a=\frac{1}{2}=0.5$$

$$b=-\frac{1}{4}=-0.25$$

$$c=\frac{1}{4}=0.25$$

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.

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

Solve these equations for b and c respectively.

$$b=3c-1 c=b+\frac{1}{2}$$

Substitute 3c-1 for b in the equation c=b+\frac{1}{2}.

$$c=3c-1+\frac{1}{2}$$

Solve c=3c-1+\frac{1}{2} for c.

$$c=\frac{1}{4}$$

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

$$b=3\times \frac{1}{4}-1$$

Calculate b from b=3\times \frac{1}{4}-1.

$$b=-\frac{1}{4}$$

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

$$a=-\left(-\frac{1}{4}\right)+\frac{1}{4}$$

Calculate a from a=-\left(-\frac{1}{4}\right)+\frac{1}{4}.

$$a=\frac{1}{2}$$

The system is now solved.

$$a=\frac{1}{2} b=-\frac{1}{4} c=\frac{1}{4}$$