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

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

Solve for b, c, a

$$b = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$c=-4$$

$$a=0$$

Short Steps Using Substitution

Solve 2b-c=1 for c.

$$c=2b-1$$

Substitute 2b-1 for c in the second and third equation.

$$4a+2b-1=-4 2a+2b-1=-4$$

Solve these equations for b and a respectively.

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

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

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

Solve a=-\left(-2a-\frac{3}{2}\right)-\frac{3}{2} for a.

$$a=0$$

Substitute 0 for a in the equation b=-2a-\frac{3}{2}.

$$b=-2\times 0-\frac{3}{2}$$

Calculate b from b=-2\times 0-\frac{3}{2}.

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

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

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

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

$$c=-4$$

The system is now solved.

$$b=-\frac{3}{2} c=-4 a=0$$