Solve {l}{1=a+b+c}{5=4a+2b+c}{12=9a+3b+c}

$$\left\{ \begin{array} { l } { 1 = a + b + c } \\ { 5 = 4 a + 2 b + c } \\ { 12 = 9 a + 3 b + c } \end{array} \right.$$

Solve for a, b, c

$$a = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

$$b=-\frac{1}{2}=-0.5$$

$$c=0$$

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.

$$5=4\left(-b+1-c\right)+2b+c 12=9\left(-b+1-c\right)+3b+c$$

Solve these equations for b and c respectively.

$$b=-\frac{1}{2}-\frac{3}{2}c c=-\frac{3}{8}-\frac{3}{4}b$$

Substitute -\frac{1}{2}-\frac{3}{2}c for b in the equation c=-\frac{3}{8}-\frac{3}{4}b.

$$c=-\frac{3}{8}-\frac{3}{4}\left(-\frac{1}{2}-\frac{3}{2}c\right)$$

Solve c=-\frac{3}{8}-\frac{3}{4}\left(-\frac{1}{2}-\frac{3}{2}c\right) for c.

$$c=0$$

Substitute 0 for c in the equation b=-\frac{1}{2}-\frac{3}{2}c.

$$b=-\frac{1}{2}-\frac{3}{2}\times 0$$

Calculate b from b=-\frac{1}{2}-\frac{3}{2}\times 0.

$$b=-\frac{1}{2}$$

Substitute -\frac{1}{2} for b and 0 for c in the equation a=-b+1-c.

$$a=-\left(-\frac{1}{2}\right)+1-0$$

Calculate a from a=-\left(-\frac{1}{2}\right)+1-0.

$$a=\frac{3}{2}$$

The system is now solved.

$$a=\frac{3}{2} b=-\frac{1}{2} c=0$$