Solve {l}{2C+3A+11=5B}{2B-3C+4A=17}{7C+6A+4=3B}

$$\left\{ \begin{array} { l } { 2 C + 3 A + 11 = 5 B } \\ { 2 B - 3 C + 4 A = 17 } \\ { 7 C + 6 A + 4 = 3 B } \end{array} \right.$$

Solve for C, A, B

$$C=-1$$

$$A=2$$

$$B=3$$

Short Steps Using Substitution

Solve 2C+3A+11=5B for C.

$$C=-\frac{3}{2}A-\frac{11}{2}+\frac{5}{2}B$$

Substitute -\frac{3}{2}A-\frac{11}{2}+\frac{5}{2}B for C in the second and third equation.

$$2B-3\left(-\frac{3}{2}A-\frac{11}{2}+\frac{5}{2}B\right)+4A=17 7\left(-\frac{3}{2}A-\frac{11}{2}+\frac{5}{2}B\right)+6A+4=3B$$

Solve these equations for A and B respectively.

$$A=\frac{1}{17}+\frac{11}{17}B B=\frac{69}{29}+\frac{9}{29}A$$

Substitute \frac{1}{17}+\frac{11}{17}B for A in the equation B=\frac{69}{29}+\frac{9}{29}A.

$$B=\frac{69}{29}+\frac{9}{29}\left(\frac{1}{17}+\frac{11}{17}B\right)$$

Solve B=\frac{69}{29}+\frac{9}{29}\left(\frac{1}{17}+\frac{11}{17}B\right) for B.

$$B=3$$

Substitute 3 for B in the equation A=\frac{1}{17}+\frac{11}{17}B.

$$A=\frac{1}{17}+\frac{11}{17}\times 3$$

Calculate A from A=\frac{1}{17}+\frac{11}{17}\times 3.

$$A=2$$

Substitute 2 for A and 3 for B in the equation C=-\frac{3}{2}A-\frac{11}{2}+\frac{5}{2}B.

$$C=-\frac{3}{2}\times 2-\frac{11}{2}+\frac{5}{2}\times 3$$

Calculate C from C=-\frac{3}{2}\times 2-\frac{11}{2}+\frac{5}{2}\times 3.

$$C=-1$$

The system is now solved.

$$C=-1 A=2 B=3$$