Solve {l}{x^2+2y^2=8}{y=m(x-2)+sqrt{2}}

$$\left. \begin{array} { l } { x ^ { 2 } + 2 y ^ { 2 } = 8 } \\ { y = m ( x - 2 ) + \sqrt { 2 } } \end{array} \right.$$

Solve for x, y

$$x=\frac{2\left(2m^{2}-\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}\text{, }y=\frac{\sqrt{2}\left(-2m|\frac{\sqrt{2}\left(\sqrt{2}m+1\right)}{2}|-\sqrt{2}m+1\right)}{2m^{2}+1}$$

$$x=\frac{2\left(2m^{2}+\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}\text{, }y=\frac{\sqrt{2}\left(2m|\frac{\sqrt{2}\left(\sqrt{2}m+1\right)}{2}|-\sqrt{2}m+1\right)}{2m^{2}+1}$$

Steps Using Substitution and Quadratic Formula

Consider the second equation. Use the distributive property to multiply m by x-2.

$$y=mx-2m+\sqrt{2}$$

Substitute mx-2m+\sqrt{2} for y in the other equation, x^{2}+2y^{2}=8.

$$x^{2}+2\left(mx-2m+\sqrt{2}\right)^{2}=8$$

Square mx-2m+\sqrt{2}.

$$x^{2}+2\left(m^{2}x^{2}+2m\left(-2m+\sqrt{2}\right)x+\left(-2m+\sqrt{2}\right)^{2}\right)=8$$

Multiply 2 times m^{2}x^{2}+2m\left(-2m+\sqrt{2}\right)x+\left(-2m+\sqrt{2}\right)^{2}.

$$x^{2}+2m^{2}x^{2}+4m\left(-2m+\sqrt{2}\right)x+2\left(-2m+\sqrt{2}\right)^{2}=8$$

Add x^{2} to 2m^{2}x^{2}.

$$\left(2m^{2}+1\right)x^{2}+4m\left(-2m+\sqrt{2}\right)x+2\left(-2m+\sqrt{2}\right)^{2}=8$$

Subtract 8 from both sides of the equation.

$$\left(2m^{2}+1\right)x^{2}+4m\left(-2m+\sqrt{2}\right)x+2\left(-2m+\sqrt{2}\right)^{2}-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+2m^{2} for a, 2\times 2m\left(-2m+\sqrt{2}\right) for b, and -4+8m^{2}-8m\sqrt{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4m\left(-2m+\sqrt{2}\right)±\sqrt{\left(4m\left(-2m+\sqrt{2}\right)\right)^{2}-4\left(2m^{2}+1\right)\left(8m^{2}-8\sqrt{2}m-4\right)}}{2\left(2m^{2}+1\right)}$$

Square 2\times 2m\left(-2m+\sqrt{2}\right).

$$x=\frac{-4m\left(-2m+\sqrt{2}\right)±\sqrt{16m^{2}\left(-2m+\sqrt{2}\right)^{2}-4\left(2m^{2}+1\right)\left(8m^{2}-8\sqrt{2}m-4\right)}}{2\left(2m^{2}+1\right)}$$

Multiply -4 times 1+2m^{2}.

$$x=\frac{-4m\left(-2m+\sqrt{2}\right)±\sqrt{16m^{2}\left(-2m+\sqrt{2}\right)^{2}+\left(-8m^{2}-4\right)\left(8m^{2}-8\sqrt{2}m-4\right)}}{2\left(2m^{2}+1\right)}$$

Multiply -4-8m^{2} times -4+8m^{2}-8m\sqrt{2}.

$$x=\frac{-4m\left(-2m+\sqrt{2}\right)±\sqrt{16m^{2}\left(-2m+\sqrt{2}\right)^{2}-64m^{4}+64\sqrt{2}m^{3}+32\sqrt{2}m+16}}{2\left(2m^{2}+1\right)}$$

Add 16m^{2}\left(-2m+\sqrt{2}\right)^{2} to 16+32m\sqrt{2}-64m^{4}+64m^{3}\sqrt{2}.

$$x=\frac{-4m\left(-2m+\sqrt{2}\right)±\sqrt{32m^{2}+32\sqrt{2}m+16}}{2\left(2m^{2}+1\right)}$$

Take the square root of 16+32m^{2}+32m\sqrt{2}.

$$x=\frac{-4m\left(-2m+\sqrt{2}\right)±4\sqrt{2m^{2}+2\sqrt{2}m+1}}{2\left(2m^{2}+1\right)}$$

Multiply 2 times 1+2m^{2}.

$$x=\frac{-4m\left(-2m+\sqrt{2}\right)±4\sqrt{2m^{2}+2\sqrt{2}m+1}}{4m^{2}+2}$$

Now solve the equation x=\frac{-4m\left(-2m+\sqrt{2}\right)±4\sqrt{2m^{2}+2\sqrt{2}m+1}}{4m^{2}+2} when ± is plus. Add -4m\left(-2m+\sqrt{2}\right) to 4\sqrt{1+2m^{2}+2m\sqrt{2}}.

$$x=\frac{-4m\left(-2m+\sqrt{2}\right)+4\sqrt{2m^{2}+2\sqrt{2}m+1}}{4m^{2}+2}$$

Divide -4m\left(-2m+\sqrt{2}\right)+4\sqrt{1+2m^{2}+2m\sqrt{2}} by 2+4m^{2}.

$$x=\frac{2\left(2m^{2}+\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}$$

Now solve the equation x=\frac{-4m\left(-2m+\sqrt{2}\right)±4\sqrt{2m^{2}+2\sqrt{2}m+1}}{4m^{2}+2} when ± is minus. Subtract 4\sqrt{1+2m^{2}+2m\sqrt{2}} from -4m\left(-2m+\sqrt{2}\right).

$$x=\frac{8m^{2}-4\sqrt{2m^{2}+2\sqrt{2}m+1}-4\sqrt{2}m}{4m^{2}+2}$$

Divide 8m^{2}-4m\sqrt{2}-4\sqrt{1+2m^{2}+2m\sqrt{2}} by 2+4m^{2}.

$$x=\frac{2\left(2m^{2}-\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}$$

There are two solutions for x: \frac{2\left(2m^{2}-m\sqrt{2}+\sqrt{2m^{2}+1+2m\sqrt{2}}\right)}{1+2m^{2}} and \frac{2\left(2m^{2}-m\sqrt{2}-\sqrt{2m^{2}+1+2m\sqrt{2}}\right)}{1+2m^{2}}. Substitute \frac{2\left(2m^{2}-m\sqrt{2}+\sqrt{2m^{2}+1+2m\sqrt{2}}\right)}{1+2m^{2}} for x in the equation y=mx-2m+\sqrt{2} to find the corresponding solution for y that satisfies both equations.

$$y=m\times \frac{2\left(2m^{2}+\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}-2m+\sqrt{2}$$

Multiply m times \frac{2\left(2m^{2}-m\sqrt{2}+\sqrt{2m^{2}+1+2m\sqrt{2}}\right)}{1+2m^{2}}.

$$y=\frac{2\left(2m^{2}+\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}m-2m+\sqrt{2}$$

Now substitute \frac{2\left(2m^{2}-m\sqrt{2}-\sqrt{2m^{2}+1+2m\sqrt{2}}\right)}{1+2m^{2}} for x in the equation y=mx-2m+\sqrt{2} and solve to find the corresponding solution for y that satisfies both equations.

$$y=m\times \frac{2\left(2m^{2}-\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}-2m+\sqrt{2}$$

Multiply m times \frac{2\left(2m^{2}-m\sqrt{2}-\sqrt{2m^{2}+1+2m\sqrt{2}}\right)}{1+2m^{2}}.

$$y=\frac{2\left(2m^{2}-\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}m-2m+\sqrt{2}$$

The system is now solved.

$$y=\frac{2\left(2m^{2}+\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}m-2m+\sqrt{2},x=\frac{2\left(2m^{2}+\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}\text{ or }y=\frac{2\left(2m^{2}-\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}m-2m+\sqrt{2},x=\frac{2\left(2m^{2}-\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}$$

Solve for x, y (complex solution)

$$\left\{\begin{matrix}x=\frac{2\left(2m^{2}-\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}\text{, }y=\frac{\sqrt{2}\left(-m\sqrt{2\left(\sqrt{2}m+1\right)^{2}}-\sqrt{2}m+1\right)}{2m^{2}+1}\text{; }x=\frac{2\left(2m^{2}+\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}\text{, }y=\frac{\sqrt{2}\left(m\sqrt{2\left(\sqrt{2}m+1\right)^{2}}-\sqrt{2}m+1\right)}{2m^{2}+1}\text{, }&m\neq -\frac{\sqrt{2}i}{2}\text{ and }m\neq \frac{\sqrt{2}i}{2}\\x=-\frac{\left(-2m+\sqrt{2}\right)^{2}-4}{2m\left(-2m+\sqrt{2}\right)}\text{, }y=\frac{2m^{2}-2\sqrt{2}m+3}{-2m+\sqrt{2}}\text{, }&m=-\frac{\sqrt{2}i}{2}\text{ or }m=\frac{\sqrt{2}i}{2}\end{matrix}\right.$$