Solve {l}{7a+10b+c+149=0}{7a-10b+c+149=0}{10a-9b+c+181=0}

$$\left\{ \begin{array} { l } { 7 a + 10 b + c + 149 = 0 } \\ { 7 a - 10 b + c + 149 = 0 } \\ { 10 a - 9 b + c + 181 = 0 } \end{array} \right.$$

Solve for a, b, c

$$a = -\frac{32}{3} = -10\frac{2}{3} \approx -10.666666667$$

$$b=0$$

$$c = -\frac{223}{3} = -74\frac{1}{3} \approx -74.333333333$$

Short Steps Using Substitution

Solve 7a+10b+c+149=0 for c.

$$c=-7a-10b-149$$

Substitute -7a-10b-149 for c in the second and third equation.

$$7a-10b-7a-10b-149+149=0 10a-9b-7a-10b-149+181=0$$

Solve these equations for b and a respectively.

$$b=0 a=\frac{19}{3}b-\frac{32}{3}$$

Substitute 0 for b in the equation a=\frac{19}{3}b-\frac{32}{3}.

$$a=\frac{19}{3}\times 0-\frac{32}{3}$$

Calculate a from a=\frac{19}{3}\times 0-\frac{32}{3}.

$$a=-\frac{32}{3}$$

Substitute 0 for b and -\frac{32}{3} for a in the equation c=-7a-10b-149.

$$c=-7\left(-\frac{32}{3}\right)-10\times 0-149$$

Calculate c from c=-7\left(-\frac{32}{3}\right)-10\times 0-149.

$$c=-\frac{223}{3}$$

The system is now solved.

$$a=-\frac{32}{3} b=0 c=-\frac{223}{3}$$