Sr Examen
kx−4y=2; x−ky=3
Solución
Respuesta rápida
$$k_{1} = \frac{- \frac{\sqrt{16 y^{2} + 8 y + 9}}{2} - \frac{3}{2}}{y}$$
=
$$- \frac{\sqrt{16 y^{2} + 8 y + 9} + 3}{2 y}$$
=
$$x_{1} = \frac{3}{2} - \frac{\sqrt{16 y^{2} + 8 y + 9}}{2}$$
=
$$\frac{3}{2} - \frac{\sqrt{16 y^{2} + 8 y + 9}}{2}$$
=
=
$$- \frac{\sqrt{16 y^{2} + 8 y + 9} + 3}{2 y}$$
=
(-1.5 - 2*(0.5625 + y^2 + 0.5*y)^0.5)/y
$$x_{1} = \frac{3}{2} - \frac{\sqrt{16 y^{2} + 8 y + 9}}{2}$$
=
$$\frac{3}{2} - \frac{\sqrt{16 y^{2} + 8 y + 9}}{2}$$
=
1.5 - 2*(0.5625 + y^2 + 0.5*y)^0.5
$$k_{2} = \frac{\frac{\sqrt{16 y^{2} + 8 y + 9}}{2} - \frac{3}{2}}{y}$$
=
$$\frac{\sqrt{16 y^{2} + 8 y + 9} - 3}{2 y}$$
=
$$x_{2} = \frac{\sqrt{16 y^{2} + 8 y + 9}}{2} + \frac{3}{2}$$
=
$$\frac{\sqrt{16 y^{2} + 8 y + 9}}{2} + \frac{3}{2}$$
=
=
$$\frac{\sqrt{16 y^{2} + 8 y + 9} - 3}{2 y}$$
=
(-1.5 + 2*(0.5625 + y^2 + 0.5*y)^0.5)/y
$$x_{2} = \frac{\sqrt{16 y^{2} + 8 y + 9}}{2} + \frac{3}{2}$$
=
$$\frac{\sqrt{16 y^{2} + 8 y + 9}}{2} + \frac{3}{2}$$
=
1.5 + 2*(0.5625 + y^2 + 0.5*y)^0.5