Solve {l}{1/15(3a+5b)+c=1/2(7b+4c-9a-8)+17}{3(7a+3b+c)=2(5a+9b+4c+4)+15}{1/5(11a+5c-30)-6c=1/8(13b+c)-a}

$$\left. \begin{array} { l } { \frac { 1 } { 15 } ( 3 a + 5 b ) + c = \frac { 1 } { 2 } ( 7 b + 4 c - 9 a - 8 ) + 17 } \\ { 3 ( 7 a + 3 b + c ) = 2 ( 5 a + 9 b + 4 c + 4 ) + 15 } \\ { \frac { 1 } { 5 } ( 11 a + 5 c - 30 ) - 6 c = \frac { 1 } { 8 } ( 13 b + c ) - a } \end{array} \right.$$

Solve for a, b, c

$$a=5$$

$$b=3$$

$$c=1$$

Short Steps Using Substitution

Solve \frac{1}{15}\left(3a+5b\right)+c=\frac{1}{2}\left(7b+4c-9a-8\right)+17 for a.

$$a=\frac{130}{47}+\frac{95}{141}b+\frac{10}{47}c$$

Substitute \frac{130}{47}+\frac{95}{141}b+\frac{10}{47}c for a in the second and third equation.

$$3\left(7\left(\frac{130}{47}+\frac{95}{141}b+\frac{10}{47}c\right)+3b+c\right)=2\left(5\left(\frac{130}{47}+\frac{95}{141}b+\frac{10}{47}c\right)+9b+4c+4\right)+15 \frac{1}{5}\left(11\left(\frac{130}{47}+\frac{95}{141}b+\frac{10}{47}c\right)+5c-30\right)-6c=\frac{1}{8}\left(13b+c\right)-\left(\frac{130}{47}+\frac{95}{141}b+\frac{10}{47}c\right)$$

Solve these equations for b and c respectively.

$$b=\frac{1047}{224}-\frac{375}{224}c c=\frac{1072}{1671}+\frac{599}{5013}b$$

Substitute \frac{1047}{224}-\frac{375}{224}c for b in the equation c=\frac{1072}{1671}+\frac{599}{5013}b.

$$c=\frac{1072}{1671}+\frac{599}{5013}\left(\frac{1047}{224}-\frac{375}{224}c\right)$$

Solve c=\frac{1072}{1671}+\frac{599}{5013}\left(\frac{1047}{224}-\frac{375}{224}c\right) for c.

$$c=1$$

Substitute 1 for c in the equation b=\frac{1047}{224}-\frac{375}{224}c.

$$b=\frac{1047}{224}-\frac{375}{224}$$

Calculate b from b=\frac{1047}{224}-\frac{375}{224}.

$$b=3$$

Substitute 3 for b and 1 for c in the equation a=\frac{130}{47}+\frac{95}{141}b+\frac{10}{47}c.

$$a=\frac{130}{47}+\frac{95}{141}\times 3+\frac{10}{47}\times 1$$

Calculate a from a=\frac{130}{47}+\frac{95}{141}\times 3+\frac{10}{47}\times 1.

$$a=5$$

The system is now solved.

$$a=5 b=3 c=1$$