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

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

Solve for a, b, c

$$a=3$$

$$b=-11$$

$$c=9$$

Short Steps Using Substitution

Solve 1=a+b+c for a.

$$a=-b+1-c$$

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

$$1=4\left(-b+1-c\right)+b -1=4\left(-b+1-c\right)+2b+c$$

Solve these equations for b and c respectively.

$$b=1-\frac{4}{3}c c=-\frac{2}{3}b+\frac{5}{3}$$

Substitute 1-\frac{4}{3}c for b in the equation c=-\frac{2}{3}b+\frac{5}{3}.

$$c=-\frac{2}{3}\left(1-\frac{4}{3}c\right)+\frac{5}{3}$$

Solve c=-\frac{2}{3}\left(1-\frac{4}{3}c\right)+\frac{5}{3} for c.

$$c=9$$

Substitute 9 for c in the equation b=1-\frac{4}{3}c.

$$b=1-\frac{4}{3}\times 9$$

Calculate b from b=1-\frac{4}{3}\times 9.

$$b=-11$$

Substitute -11 for b and 9 for c in the equation a=-b+1-c.

$$a=-\left(-11\right)+1-9$$

Calculate a from a=-\left(-11\right)+1-9.

$$a=3$$

The system is now solved.

$$a=3 b=-11 c=9$$