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

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

Solve for a, b, c

$$a=1$$

$$b=-1$$

$$c=2$$

Short Steps Using Substitution

Solve a+b+c=2 for a.

$$a=-b-c+2$$

Substitute -b-c+2 for a in the second and third equation.

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

Solve these equations for b and c respectively.

$$b=-\frac{5}{4}c+\frac{3}{2} c=\frac{5}{2}+\frac{1}{2}b$$

Substitute -\frac{5}{4}c+\frac{3}{2} for b in the equation c=\frac{5}{2}+\frac{1}{2}b.

$$c=\frac{5}{2}+\frac{1}{2}\left(-\frac{5}{4}c+\frac{3}{2}\right)$$

Solve c=\frac{5}{2}+\frac{1}{2}\left(-\frac{5}{4}c+\frac{3}{2}\right) for c.

$$c=2$$

Substitute 2 for c in the equation b=-\frac{5}{4}c+\frac{3}{2}.

$$b=-\frac{5}{4}\times 2+\frac{3}{2}$$

Calculate b from b=-\frac{5}{4}\times 2+\frac{3}{2}.

$$b=-1$$

Substitute -1 for b and 2 for c in the equation a=-b-c+2.

$$a=-\left(-1\right)-2+2$$

Calculate a from a=-\left(-1\right)-2+2.

$$a=1$$

The system is now solved.

$$a=1 b=-1 c=2$$