Solve {l}{a+2b+3c=12}{3a+2b+5c=22}{2a+4b+4c=18}

$$\left. \begin{array} { l } { a + 2 b + 3 c = 12 } \\ { 3 a + 2 b + 5 c = 22 } \\ { 2 a + 4 b + 4 c = 18 } \end{array} \right.$$

Solve for a, b, c

$$a=2$$

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

$$c=3$$

Short Steps Using Substitution

Solve a+2b+3c=12 for a.

$$a=-2b-3c+12$$

Substitute -2b-3c+12 for a in the second and third equation.

$$3\left(-2b-3c+12\right)+2b+5c=22 2\left(-2b-3c+12\right)+4b+4c=18$$

Solve these equations for b and c respectively.

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

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

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

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

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

Substitute \frac{1}{2} for b and 3 for c in the equation a=-2b-3c+12.

$$a=-2\times \frac{1}{2}-3\times 3+12$$

Calculate a from a=-2\times \frac{1}{2}-3\times 3+12.

$$a=2$$

The system is now solved.

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