Solve {l}{3a+4b=7}{5a-2b+7c=8}{2a+3b+c=9}

$$\left\{ \begin{array} { l } { 3 a + 4 b = 7 } \\ { 5 a - 2 b + 7 c = 8 } \\ { 2 a + 3 b + c = 9 } \end{array} \right.$$

Solve for a, b, c

$$a = -\frac{59}{33} = -1\frac{26}{33} \approx -1.787878788$$

$$b = \frac{34}{11} = 3\frac{1}{11} \approx 3.090909091$$

$$c = \frac{109}{33} = 3\frac{10}{33} \approx 3.303030303$$

Short Steps Using Substitution

Reorder the equations.

$$2a+3b+c=9 5a-2b+7c=8 3a+4b=7$$

Solve 2a+3b+c=9 for c.

$$c=-2a-3b+9$$

Substitute -2a-3b+9 for c in the equation 5a-2b+7c=8.

$$5a-2b+7\left(-2a-3b+9\right)=8$$

Solve the second equation for b and the third equation for a.

$$b=-\frac{9}{23}a+\frac{55}{23} a=-\frac{4}{3}b+\frac{7}{3}$$

Substitute -\frac{9}{23}a+\frac{55}{23} for b in the equation a=-\frac{4}{3}b+\frac{7}{3}.

$$a=-\frac{4}{3}\left(-\frac{9}{23}a+\frac{55}{23}\right)+\frac{7}{3}$$

Solve a=-\frac{4}{3}\left(-\frac{9}{23}a+\frac{55}{23}\right)+\frac{7}{3} for a.

$$a=-\frac{59}{33}$$

Substitute -\frac{59}{33} for a in the equation b=-\frac{9}{23}a+\frac{55}{23}.

$$b=-\frac{9}{23}\left(-\frac{59}{33}\right)+\frac{55}{23}$$

Calculate b from b=-\frac{9}{23}\left(-\frac{59}{33}\right)+\frac{55}{23}.

$$b=\frac{34}{11}$$

Substitute \frac{34}{11} for b and -\frac{59}{33} for a in the equation c=-2a-3b+9.

$$c=-2\left(-\frac{59}{33}\right)-3\times \frac{34}{11}+9$$

Calculate c from c=-2\left(-\frac{59}{33}\right)-3\times \frac{34}{11}+9.

$$c=\frac{109}{33}$$

The system is now solved.

$$a=-\frac{59}{33} b=\frac{34}{11} c=\frac{109}{33}$$