System of equations: Solve the system of equation
{xy=8}
{x2+y2=20}
Formula:
\(\begin{array}{|rcll|} \hline (x+y)^2 &=& x^2+2xy+y^2 \\ x^2+y^2 &=& (x+y)^2-2xy \quad & | \quad xy = 8 \\ x^2+y^2 &=& (x+y)^2 - 8 \\ \hline \end{array}\)
1.
\(\begin{array}{|rcll|} \hline x^2+y^2 &=& 20 \quad & | \quad x^2+y^2 = (x+y)^2-16 \\ (x+y)^2-16 &=& 20 \\ (x+y)^2 &=& 20+16 \\ (x+y)^2 &=& 36 \\ \mathbf{x+y} &\mathbf{=}& \mathbf{\pm 6} \\ \hline \end{array}\)
Two Systems of equation now:
\(\begin{array}{|lrcl|lrcl|} \hline (1)& xy &=& 8 &(1) & xy &=& 8 \\ &\mathbf{y} &\mathbf{=}& \mathbf{\frac{8}{x}} & & \mathbf{y} &\mathbf{=}& \mathbf{\frac{8}{x}} \\\\ (2)& x+y &=& 6 &(2)& x+y &=& -6 \\ & x+\frac{8}{x} &=& 6 & & x+\frac{8}{x} &=& -6 \\ & x^2+8 &=& 6x & & x^2+8 &=& -6x \\ & x^2-6x+8 &=& 0 & & x^2+6x+8 &=& 0 \\ & x_{1,2}&=& \frac{6\pm \sqrt{36-4\cdot 8} }{2} & & x_{3,4}&=& \frac{-6\pm \sqrt{36-4\cdot 8} }{2} \\ & x_{1,2}&=& \frac{6\pm 4 }{2} & & x_{3,4}&=& \frac{-6\pm 4 }{2} \\ & x_{1,2}&=& 3\pm 1 & & x_{3,4}&=& -3\pm 1 \\\\ & \mathbf{x_1 =4} && \mathbf{x_2 = 2} & & \mathbf{x_3 = -2} && \mathbf{x_4 =-4} \\ & \mathbf{y_1} =\frac{8}{4}\mathbf{=2} && \mathbf{y_2} = \frac{8}{2}\mathbf{=4} & & \mathbf{y_3} = \frac{8}{-2}\mathbf{=-4} && \mathbf{y_4} = \frac{8}{-4}\mathbf{=-2} \\ \hline \end{array} \)