Solve {l}{-4=a+b+c}{20=9a-3b+c}{-12=a-b+c}

$$\left\{ \begin{array} { l } { - 4 = a + b + c } \\ { 20 = 9 a - 3 b + c } \\ { - 12 = a - b + c } \end{array} \right.$$

Solve for a, b, c

$$a=5$$

$$b=4$$

$$c=-13$$

Short Steps Using Substitution

Solve -4=a+b+c for a.

$$a=-b-4-c$$

Substitute -b-4-c for a in the second and third equation.

$$20=9\left(-b-4-c\right)-3b+c -12=-b-4-c-b+c$$

Solve these equations for c and b respectively.

$$c=-7-\frac{3}{2}b b=4$$

Substitute 4 for b in the equation c=-7-\frac{3}{2}b.

$$c=-7-\frac{3}{2}\times 4$$

Calculate c from c=-7-\frac{3}{2}\times 4.

$$c=-13$$

Substitute -13 for c and 4 for b in the equation a=-b-4-c.

$$a=-4-4-\left(-13\right)$$

Calculate a from a=-4-4-\left(-13\right).

$$a=5$$

The system is now solved.

$$a=5 b=4 c=-13$$