Solve {l}{a+b+c=7}{4a+2b+c=12}{9a+3b+c=18}

$$\left\{ \begin{array} { l } { a + b + c = 7 } \\ { 4 a + 2 b + c = 12 } \\ { 9 a + 3 b + c = 18 } \end{array} \right.$$

Solve for a, b, c

$$a=\frac{1}{2}=0.5$$

$$b = \frac{7}{2} = 3\frac{1}{2} = 3.5$$

$$c=3$$

Short Steps Using Substitution

Solve a+b+c=7 for a.

$$a=-b-c+7$$

Substitute -b-c+7 for a in the second and third equation.

$$4\left(-b-c+7\right)+2b+c=12 9\left(-b-c+7\right)+3b+c=18$$

Solve these equations for b and c respectively.

$$b=-\frac{3}{2}c+8 c=\frac{45}{8}-\frac{3}{4}b$$

Substitute -\frac{3}{2}c+8 for b in the equation c=\frac{45}{8}-\frac{3}{4}b.

$$c=\frac{45}{8}-\frac{3}{4}\left(-\frac{3}{2}c+8\right)$$

Solve c=\frac{45}{8}-\frac{3}{4}\left(-\frac{3}{2}c+8\right) for c.

$$c=3$$

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

$$b=-\frac{3}{2}\times 3+8$$

Calculate b from b=-\frac{3}{2}\times 3+8.

$$b=\frac{7}{2}$$

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

$$a=-\frac{7}{2}-3+7$$

Calculate a from a=-\frac{7}{2}-3+7.

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

The system is now solved.

$$a=\frac{1}{2} b=\frac{7}{2} c=3$$