Solve {l}{a+b-c+1=0}{2a-b+c-4=0}{b+c-1=0}

$$\left. \begin{array} { l } { a + b - c + 1 = 0 } \\ { 2 a - b + c - 4 = 0 } \\ { b + c - 1 = 0 } \end{array} \right.$$

Solve for a, b, c

$$a=1$$

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

$$c = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Short Steps Using Substitution

Solve a+b-c+1=0 for a.

$$a=-b+c-1$$

Substitute -b+c-1 for a in the equation 2a-b+c-4=0.

$$2\left(-b+c-1\right)-b+c-4=0$$

Solve the second equation for b and the third equation for c.

$$b=c-2 c=-b+1$$

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

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

Solve c=-\left(c-2\right)+1 for c.

$$c=\frac{3}{2}$$

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

$$b=\frac{3}{2}-2$$

Calculate b from b=\frac{3}{2}-2.

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

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

$$a=-\left(-\frac{1}{2}\right)+\frac{3}{2}-1$$

Calculate a from a=-\left(-\frac{1}{2}\right)+\frac{3}{2}-1.

$$a=1$$

The system is now solved.

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