Solve {l}{6=a+b+c}{12=4a+2b+c}{20=9a+3b+c}

$$\left. \begin{array}{l}{ 6 = a + b + c }\\{ 12 = 4 a + 2 b + c }\\{ 20 = 9 a + 3 b + c }\end{array} \right.$$

Solve for a, b, c

$$a=1$$

$$b=3$$

$$c=2$$

Short Steps Using Substitution

Solve 6=a+b+c for a.

$$a=-b+6-c$$

Substitute -b+6-c for a in the second and third equation.

$$12=4\left(-b+6-c\right)+2b+c 20=9\left(-b+6-c\right)+3b+c$$

Solve these equations for b and c respectively.

$$b=6-\frac{3}{2}c c=\frac{17}{4}-\frac{3}{4}b$$

Substitute 6-\frac{3}{2}c for b in the equation c=\frac{17}{4}-\frac{3}{4}b.

$$c=\frac{17}{4}-\frac{3}{4}\left(6-\frac{3}{2}c\right)$$

Solve c=\frac{17}{4}-\frac{3}{4}\left(6-\frac{3}{2}c\right) for c.

$$c=2$$

Substitute 2 for c in the equation b=6-\frac{3}{2}c.

$$b=6-\frac{3}{2}\times 2$$

Calculate b from b=6-\frac{3}{2}\times 2.

$$b=3$$

Substitute 3 for b and 2 for c in the equation a=-b+6-c.

$$a=-3+6-2$$

Calculate a from a=-3+6-2.

$$a=1$$

The system is now solved.

$$a=1 b=3 c=2$$