Solve {l}{6=9a+(-3)b+c}{-1=4a-2b+c}{-3=c}

$$\left\{ \begin{array} { l } { 6 = 9 a + ( - 3 ) b + c } \\ { - 1 = 4 a - 2 b + c } \\ { - 3 = c } \end{array} \right.$$

Solve for a, b, c

$$a=2$$

$$b=3$$

$$c=-3$$

Short Steps Using Substitution

Reorder the equations.

$$-3=c -1=4a-2b+c 6=9a-3b+c$$

Solve -3=c for c.

$$c=-3$$

Substitute -3 for c in the second and third equation.

$$-1=4a-2b-3 6=9a-3b-3$$

Solve these equations for b and a respectively.

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

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

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

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

$$a=2$$

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

$$b=-1+2\times 2$$

Calculate b from b=-1+2\times 2.

$$b=3$$

The system is now solved.

$$a=2 b=3 c=-3$$