Solve {c}{8A+5C-5D=0}{C-A=2}{11D-2A-3C=270}

$$\left. \begin{array} { c } { 8 A + 5 C - 5 D = 0 } \\ { C - A = 2 } \\ { 11 D - 2 A - 3 C = 270 } \end{array} \right.$$

Solve for A, C, D

$$A = \frac{635}{59} = 10\frac{45}{59} \approx 10.762711864$$

$$C = \frac{753}{59} = 12\frac{45}{59} \approx 12.762711864$$

$$D = \frac{1769}{59} = 29\frac{58}{59} \approx 29.983050847$$

Short Steps Using Substitution

Reorder the equations.

$$C-A=2 8A+5C-5D=0 11D-2A-3C=270$$

Solve C-A=2 for A.

$$A=C-2$$

Substitute C-2 for A in the second and third equation.

$$8\left(C-2\right)+5C-5D=0 11D-2\left(C-2\right)-3C=270$$

Solve these equations for C and D respectively.

$$C=\frac{16}{13}+\frac{5}{13}D D=\frac{5}{11}C+\frac{266}{11}$$

Substitute \frac{16}{13}+\frac{5}{13}D for C in the equation D=\frac{5}{11}C+\frac{266}{11}.

$$D=\frac{5}{11}\left(\frac{16}{13}+\frac{5}{13}D\right)+\frac{266}{11}$$

Solve D=\frac{5}{11}\left(\frac{16}{13}+\frac{5}{13}D\right)+\frac{266}{11} for D.

$$D=\frac{1769}{59}$$

Substitute \frac{1769}{59} for D in the equation C=\frac{16}{13}+\frac{5}{13}D.

$$C=\frac{16}{13}+\frac{5}{13}\times \frac{1769}{59}$$

Calculate C from C=\frac{16}{13}+\frac{5}{13}\times \frac{1769}{59}.

$$C=\frac{753}{59}$$

Substitute \frac{753}{59} for C in the equation A=C-2.

$$A=\frac{753}{59}-2$$

Calculate A from A=\frac{753}{59}-2.

$$A=\frac{635}{59}$$

The system is now solved.

$$A=\frac{635}{59} C=\frac{753}{59} D=\frac{1769}{59}$$