help Me please.

xy  =  8      Divide both sides of this equation by  y .

x  =  8/y     Use this value for  x  in the second given equation.

x² + y²  =  20

(8/y)² + y²   =   20

64/y² + y²   =   20           Multiply through by  y²

64 + y⁴  =  20y²              Subtract  20y²  from both sides.

y⁴ - 20y² + 64  =  0         We can rewrite this as...

(y²)² - 20y² + 64  =  0     Factor like this:   u² - 20u + 64  =   (u - 4)(u - 16)

(y² - 4)(y² - 16)  =  0       Factor each difference of squares.

(y + 2)(y - 2)(y + 4)(y - 4)  =  0      Set each factor equal to zero.

y + 2  =  0     or     y - 2  =  0     or     y + 4  =  0     or     y - 4  =  0

y  =  -2          or     y  =  2          or     y  =  -4          or     y  =  4

Find  x  at each of these values for  y .

When  y  =  -2  ,     x  =  8 / y  =  8 / -2  =  -4

When  y  =  2  ,     x  =  8 / y  =  8 / 2  =  4

When  y  =  -4  ,     x  =  8 / y  =  8 / -4  =  -2

When  y  =  4  ,     x  =  8 / y  =  8 / 4  =  2

The solution set is   { (-2, -4), (2, 4), (-4, -2), (4, 2) }

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:

\(\begin{array}{|lrcl|lrcl|} \hline (1)& xy &=& 8 &(1) & xy &=& 8 \\ &\mathbf{y} &\mathbf{=}& \mathbf{\frac{8}{x}} & & \mathbf{y} &\mathbf{=}& \mathbf{\frac{8}{x}} \\\\ (2)& x+y &=& 6 &(2)& x+y &=& -6 \\ & x+\frac{8}{x} &=& 6 & & x+\frac{8}{x} &=& -6 \\ & x^2+8 &=& 6x & & x^2+8 &=& -6x \\ & x^2-6x+8 &=& 0 & & x^2+6x+8 &=& 0 \\ & x_{1,2}&=& \frac{6\pm \sqrt{36-4\cdot 8} }{2} & & x_{3,4}&=& \frac{-6\pm \sqrt{36-4\cdot 8} }{2} \\ & x_{1,2}&=& \frac{6\pm 4 }{2} & & x_{3,4}&=& \frac{-6\pm 4 }{2} \\ & x_{1,2}&=& 3\pm 1 & & x_{3,4}&=& -3\pm 1 \\\\ & \mathbf{x_1 =4} && \mathbf{x_2 = 2} & & \mathbf{x_3 = -2} && \mathbf{x_4 =-4} \\ & \mathbf{y_1} =\frac{8}{4}\mathbf{=2} && \mathbf{y_2} = \frac{8}{2}\mathbf{=4} & & \mathbf{y_3} = \frac{8}{-2}\mathbf{=-4} && \mathbf{y_4} = \frac{8}{-4}\mathbf{=-2} \\ \hline \end{array} \)