Solve {l}{-2a+b+3c=-2}{3a-2b=13}{a+b+2c=9}

$$\left. \begin{array} { l } { - 2 a + b + 3 c = - 2 } \\ { 3 a - 2 b = 13 } \\ { a + b + 2 c = 9 } \end{array} \right.$$

Solve for a, b, c

$$a = \frac{75}{17} = 4\frac{7}{17} \approx 4.411764706$$

$$b=\frac{2}{17}\approx 0.117647059$$

$$c = \frac{38}{17} = 2\frac{4}{17} \approx 2.235294118$$

Short Steps Using Substitution

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

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

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

$$3a-2\left(2a-3c-2\right)=13 a+2a-3c-2+2c=9$$

Solve these equations for a and c respectively.

$$a=6c-9 c=3a-11$$

Substitute 6c-9 for a in the equation c=3a-11.

$$c=3\left(6c-9\right)-11$$

Solve c=3\left(6c-9\right)-11 for c.

$$c=\frac{38}{17}$$

Substitute \frac{38}{17} for c in the equation a=6c-9.

$$a=6\times \frac{38}{17}-9$$

Calculate a from a=6\times \frac{38}{17}-9.

$$a=\frac{75}{17}$$

Substitute \frac{75}{17} for a and \frac{38}{17} for c in the equation b=2a-3c-2.

$$b=2\times \frac{75}{17}-3\times \frac{38}{17}-2$$

Calculate b from b=2\times \frac{75}{17}-3\times \frac{38}{17}-2.

$$b=\frac{2}{17}$$

The system is now solved.

$$a=\frac{75}{17} b=\frac{2}{17} c=\frac{38}{17}$$