Solve {l}{3a+2b+c=4}{2a+b+3c=5}{5a+4b+7c=12}

$$\left. \begin{array} { l } { 3 a + 2 b + c = 4 } \\ { 2 a + b + 3 c = 5 } \\ { 5 a + 4 b + 7 c = 12 } \end{array} \right.$$

Solve for a, b, c

$$a=1$$

$$b=0$$

$$c=1$$

Short Steps Using Substitution

Solve 3a+2b+c=4 for c.

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

Substitute -3a-2b+4 for c in the second and third equation.

$$2a+b+3\left(-3a-2b+4\right)=5 5a+4b+7\left(-3a-2b+4\right)=12$$

Solve these equations for b and a respectively.

$$b=\frac{7}{5}-\frac{7}{5}a a=1-\frac{5}{8}b$$

Substitute \frac{7}{5}-\frac{7}{5}a for b in the equation a=1-\frac{5}{8}b.

$$a=1-\frac{5}{8}\left(\frac{7}{5}-\frac{7}{5}a\right)$$

Solve a=1-\frac{5}{8}\left(\frac{7}{5}-\frac{7}{5}a\right) for a.

$$a=1$$

Substitute 1 for a in the equation b=\frac{7}{5}-\frac{7}{5}a.

$$b=\frac{7}{5}-\frac{7}{5}$$

Calculate b from b=\frac{7}{5}-\frac{7}{5}.

$$b=0$$

Substitute 0 for b and 1 for a in the equation c=-3a-2b+4.

$$c=-3-2\times 0+4$$

Calculate c from c=-3-2\times 0+4.

$$c=1$$

The system is now solved.

$$a=1 b=0 c=1$$