Solve {l}{4a+3b+c=0}{a+b+c=2}{c=3}

$$\left\{ \begin{array} { l } { 4 a + 3 b + c = 0 } \\ { a + b + c = 2 } \\ { c = 3 } \end{array} \right.$$

Solve for a, b, c

$$a=0$$

$$b=-1$$

$$c=3$$

Short Steps Using Substitution

Reorder the equations.

$$c=3 a+b+c=2 4a+3b+c=0$$

Substitute 3 for c in the second and third equation.

$$a+b+3=2 4a+3b+3=0$$

Solve these equations for b and a respectively.

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

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

$$a=-\frac{3}{4}-\frac{3}{4}\left(-a-1\right)$$

Solve a=-\frac{3}{4}-\frac{3}{4}\left(-a-1\right) for a.

$$a=0$$

Substitute 0 for a in the equation b=-a-1.

$$b=-0-1$$

Calculate b from b=-0-1.

$$b=-1$$

The system is now solved.

$$a=0 b=-1 c=3$$