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

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

Solve for a, b, c

$$a = -\frac{26}{5} = -5\frac{1}{5} = -5.2$$

$$b = \frac{19}{5} = 3\frac{4}{5} = 3.8$$

$$c = \frac{22}{5} = 4\frac{2}{5} = 4.4$$

Short Steps Using Substitution

Reorder the equations.

$$a+b+c=3 3a+4c=2 2a+3b=1$$

Solve a+b+c=3 for a.

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

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

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

Solve these equations for b and c respectively.

$$b=\frac{7}{3}+\frac{1}{3}c c=\frac{5}{2}+\frac{1}{2}b$$

Substitute \frac{7}{3}+\frac{1}{3}c for b in the equation c=\frac{5}{2}+\frac{1}{2}b.

$$c=\frac{5}{2}+\frac{1}{2}\left(\frac{7}{3}+\frac{1}{3}c\right)$$

Solve c=\frac{5}{2}+\frac{1}{2}\left(\frac{7}{3}+\frac{1}{3}c\right) for c.

$$c=\frac{22}{5}$$

Substitute \frac{22}{5} for c in the equation b=\frac{7}{3}+\frac{1}{3}c.

$$b=\frac{7}{3}+\frac{1}{3}\times \frac{22}{5}$$

Calculate b from b=\frac{7}{3}+\frac{1}{3}\times \frac{22}{5}.

$$b=\frac{19}{5}$$

Substitute \frac{19}{5} for b and \frac{22}{5} for c in the equation a=-b-c+3.

$$a=-\frac{19}{5}-\frac{22}{5}+3$$

Calculate a from a=-\frac{19}{5}-\frac{22}{5}+3.

$$a=-\frac{26}{5}$$

The system is now solved.

$$a=-\frac{26}{5} b=\frac{19}{5} c=\frac{22}{5}$$