Solve {l}{4+0+2A+0+C=0}{4+9+2A+3B+C=0}{1+9+A+3B+C=0}

$$\left\{ \begin{array} { l } { 4 + 0 + 2 A + 0 + C = 0 } \\ { 4 + 9 + 2 A + 3 B + C = 0 } \\ { 1 + 9 + A + 3 B + C = 0 } \end{array} \right.$$

Solve for A, C, B

$$A=-3$$

$$C=2$$

$$B=-3$$

Short Steps Using Substitution

Solve 4+0+2A+0+C=0 for C.

$$C=-4-2A$$

Substitute -4-2A for C in the second and third equation.

$$4+9+2A+3B-4-2A=0 1+9+A+3B-4-2A=0$$

Solve these equations for B and A respectively.

$$B=-3 A=6+3B$$

Substitute -3 for B in the equation A=6+3B.

$$A=6+3\left(-3\right)$$

Calculate A from A=6+3\left(-3\right).

$$A=-3$$

Substitute -3 for A in the equation C=-4-2A.

$$C=-4-2\left(-3\right)$$

Calculate C from C=-4-2\left(-3\right).

$$C=2$$

The system is now solved.

$$A=-3 C=2 B=-3$$