{l}{x^2+y^2=8}{z=x+y}{a=z}{b=a}{text{Solvefor}ctext{where}}{c=b} ઉકેલો

$$\left. \begin{array} { l } { x ^ {2} + y ^ {2} = 8 }\\ { z = x + y }\\ { a = z }\\ { b = a }\\ { \text{Solve for } c \text{ where} } \\ { c = b } \end{array} \right.$$

x, y, z માટે ઉકેલો

$$\left\{\begin{matrix}x=-\frac{\sqrt{2\left(-b\sqrt{16-b^{2}}+8\right)}}{2}\text{, }y=\frac{\sqrt{16-b^{2}}+b}{2}\text{, }z=b\text{, }&c=b\text{ and }a=b\text{ and }b\leq 2\sqrt{2}\text{ and }|b|\leq 4\\x=-\frac{\sqrt{2\left(b\sqrt{16-b^{2}}+8\right)}}{2}\text{, }y=\frac{-\sqrt{16-b^{2}}+b}{2}\text{, }z=b\text{, }&c=b\text{ and }a=b\text{ and }b\geq -4\text{ and }b\leq -2\sqrt{2}\\x=\frac{\sqrt{2\left(-b\sqrt{16-b^{2}}+8\right)}}{2}\text{, }y=\frac{\sqrt{16-b^{2}}+b}{2}\text{, }z=b\text{, }&c=b\text{ and }a=b\text{ and }b\geq 2\sqrt{2}\text{ and }b\leq 4\\x=\frac{\sqrt{2\left(b\sqrt{16-b^{2}}+8\right)}}{2}\text{, }y=\frac{-\sqrt{16-b^{2}}+b}{2}\text{, }z=b\text{, }&c=b\text{ and }a=b\text{ and }b\geq -2\sqrt{2}\text{ and }|b|\leq 4\end{matrix}\right.$$