Solve {l}{3ac-a^2=ab+3bc=32}{a+b=4}{3c+b=}

$$\left. \begin{array} { l } { 3 a c - a ^ { 2 } = a b + 3 b c = 32 } \\ { a + b = 4 } \\ { 3 c + b = } \end{array} \right.$$

Solve for a, c, b (complex solution)

$$a=\frac{\sqrt{b^{2}-18bc+9c^{2}}+3c-b}{2}\text{, }c\in \mathrm{C}\text{, }b\in \mathrm{C}$$

$$a=\frac{-\sqrt{b^{2}-18bc+9c^{2}}+3c-b}{2}\text{, }c\in \mathrm{C}\text{, }b\in \mathrm{C}$$

Solve for a, c, b

$$\left\{\begin{matrix}\\a=\frac{\sqrt{b^{2}-18bc+9c^{2}}+3c-b}{2}\text{, }c=-\frac{2\sqrt{2}b}{3}+b\text{, }b\in \mathrm{R}\text{; }a=\frac{\sqrt{b^{2}-18bc+9c^{2}}+3c-b}{2}\text{, }c=\frac{2\sqrt{2}b}{3}+b\text{, }b\in \mathrm{R}\text{; }a=\frac{-\sqrt{b^{2}-18bc+9c^{2}}+3c-b}{2}\text{, }c=-\frac{2\sqrt{2}b}{3}+b\text{, }b\in \mathrm{R}\text{; }a=\frac{-\sqrt{b^{2}-18bc+9c^{2}}+3c-b}{2}\text{, }c=\frac{2\sqrt{2}b}{3}+b\text{, }b\in \mathrm{R}\text{, }&\text{unconditionally}\\a=\frac{\sqrt{b^{2}-18bc+9c^{2}}+3c-b}{2}\text{, }c\geq -\frac{2\sqrt{2}b}{3}+b\text{, }b\in \mathrm{R}\text{; }a=\frac{-\sqrt{b^{2}-18bc+9c^{2}}+3c-b}{2}\text{, }c\geq -\frac{2\sqrt{2}b}{3}+b\text{, }b\in \mathrm{R}\text{; }a=\frac{\sqrt{b^{2}-18bc+9c^{2}}+3c-b}{2}\text{, }c\leq \frac{2\sqrt{2}b}{3}+b\text{, }b\in \mathrm{R}\text{; }a=\frac{-\sqrt{b^{2}-18bc+9c^{2}}+3c-b}{2}\text{, }c\leq \frac{2\sqrt{2}b}{3}+b\text{, }b\in \mathrm{R}\text{, }&b<0\\a=\frac{\sqrt{b^{2}-18bc+9c^{2}}+3c-b}{2}\text{, }c\geq \frac{2\sqrt{2}b}{3}+b\text{, }b\in \mathrm{R}\text{; }a=\frac{-\sqrt{b^{2}-18bc+9c^{2}}+3c-b}{2}\text{, }c\geq \frac{2\sqrt{2}b}{3}+b\text{, }b\in \mathrm{R}\text{; }a=\frac{\sqrt{b^{2}-18bc+9c^{2}}+3c-b}{2}\text{, }c\leq -\frac{2\sqrt{2}b}{3}+b\text{, }b\in \mathrm{R}\text{; }a=\frac{-\sqrt{b^{2}-18bc+9c^{2}}+3c-b}{2}\text{, }c\leq -\frac{2\sqrt{2}b}{3}+b\text{, }b\in \mathrm{R}\text{, }&b\geq 0\end{matrix}\right.$$