Solve {l}{x^2/4+y^2/2=1}{x=my+1}

$$\left\{ \begin{array} { l } { \frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 2 } = 1 } \\ { x = m y + 1 } \end{array} \right.$$

Solve for x, y

$$x=\frac{\sqrt{2}\left(-m\sqrt{2m^{2}+3}+\sqrt{2}\right)}{m^{2}+2}\text{, }y=-\frac{\sqrt{2\left(2m^{2}+3\right)}+m}{m^{2}+2}$$

$$x=\frac{\sqrt{2}\left(m\sqrt{2m^{2}+3}+\sqrt{2}\right)}{m^{2}+2}\text{, }y=\frac{\sqrt{2\left(2m^{2}+3\right)}-m}{m^{2}+2}$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Multiply both sides of the equation by 4, the least common multiple of 4,2.

$$x^{2}+2y^{2}=4$$

Consider the second equation. Subtract my from both sides.

$$x-my=1$$

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

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

Solve x+\left(-m\right)y=1 for x by isolating x on the left hand side of the equal sign.

$$x+\left(-m\right)y=1$$

Subtract \left(-m\right)y from both sides of the equation.

$$x=my+1$$

Substitute my+1 for x in the other equation, 2y^{2}+x^{2}=4.

$$2y^{2}+\left(my+1\right)^{2}=4$$

Square my+1.

$$2y^{2}+m^{2}y^{2}+2my+1=4$$

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

$$\left(m^{2}+2\right)y^{2}+2my+1=4$$

Subtract 4 from both sides of the equation.

$$\left(m^{2}+2\right)y^{2}+2my-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2+1m^{2} for a, 1\times 1\times 2m for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-2m±\sqrt{\left(2m\right)^{2}-4\left(m^{2}+2\right)\left(-3\right)}}{2\left(m^{2}+2\right)}$$

Square 1\times 1\times 2m.

$$y=\frac{-2m±\sqrt{4m^{2}-4\left(m^{2}+2\right)\left(-3\right)}}{2\left(m^{2}+2\right)}$$

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

$$y=\frac{-2m±\sqrt{4m^{2}+\left(-4m^{2}-8\right)\left(-3\right)}}{2\left(m^{2}+2\right)}$$

Multiply -8-4m^{2} times -3.

$$y=\frac{-2m±\sqrt{4m^{2}+12m^{2}+24}}{2\left(m^{2}+2\right)}$$

Add 4m^{2} to 24+12m^{2}.

$$y=\frac{-2m±\sqrt{16m^{2}+24}}{2\left(m^{2}+2\right)}$$

Take the square root of 24+16m^{2}.

$$y=\frac{-2m±2\sqrt{4m^{2}+6}}{2\left(m^{2}+2\right)}$$

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

$$y=\frac{-2m±2\sqrt{4m^{2}+6}}{2m^{2}+4}$$

Now solve the equation y=\frac{-2m±2\sqrt{4m^{2}+6}}{2m^{2}+4} when ± is plus. Add -2m to 2\sqrt{6+4m^{2}}.

$$y=\frac{2\sqrt{4m^{2}+6}-2m}{2m^{2}+4}$$

Divide -2m+2\sqrt{6+4m^{2}} by 4+2m^{2}.

$$y=\frac{\sqrt{4m^{2}+6}-m}{m^{2}+2}$$

Now solve the equation y=\frac{-2m±2\sqrt{4m^{2}+6}}{2m^{2}+4} when ± is minus. Subtract 2\sqrt{6+4m^{2}} from -2m.

$$y=\frac{-2\sqrt{4m^{2}+6}-2m}{2m^{2}+4}$$

Divide -2m-2\sqrt{6+4m^{2}} by 4+2m^{2}.

$$y=-\frac{\sqrt{4m^{2}+6}+m}{m^{2}+2}$$

There are two solutions for y: \frac{-m+\sqrt{6+4m^{2}}}{2+m^{2}} and -\frac{m+\sqrt{6+4m^{2}}}{2+m^{2}}. Substitute \frac{-m+\sqrt{6+4m^{2}}}{2+m^{2}} for y in the equation x=my+1 to find the corresponding solution for x that satisfies both equations.

$$x=m\times \frac{\sqrt{4m^{2}+6}-m}{m^{2}+2}+1$$

Multiply m times \frac{-m+\sqrt{6+4m^{2}}}{2+m^{2}}.

$$x=\frac{\sqrt{4m^{2}+6}-m}{m^{2}+2}m+1$$

Add m\times \frac{-m+\sqrt{6+4m^{2}}}{2+m^{2}} to 1.

$$x=1+\frac{\sqrt{4m^{2}+6}-m}{m^{2}+2}m$$

Now substitute -\frac{m+\sqrt{6+4m^{2}}}{2+m^{2}} for y in the equation x=my+1 and solve to find the corresponding solution for x that satisfies both equations.

$$x=m\left(-\frac{\sqrt{4m^{2}+6}+m}{m^{2}+2}\right)+1$$

Multiply m times -\frac{m+\sqrt{6+4m^{2}}}{2+m^{2}}.

$$x=\left(-\frac{\sqrt{4m^{2}+6}+m}{m^{2}+2}\right)m+1$$

Add m\left(-\frac{m+\sqrt{6+4m^{2}}}{2+m^{2}}\right) to 1.

$$x=1+\left(-\frac{\sqrt{4m^{2}+6}+m}{m^{2}+2}\right)m$$

The system is now solved.

$$x=1+\frac{\sqrt{4m^{2}+6}-m}{m^{2}+2}m,y=\frac{\sqrt{4m^{2}+6}-m}{m^{2}+2}\text{ or }x=1+\left(-\frac{\sqrt{4m^{2}+6}+m}{m^{2}+2}\right)m,y=-\frac{\sqrt{4m^{2}+6}+m}{m^{2}+2}$$

Solve for x, y (complex solution)

$$\left\{\begin{matrix}x=\frac{\sqrt{2}\left(-m\sqrt{2m^{2}+3}+\sqrt{2}\right)}{m^{2}+2}\text{, }y=-\frac{\sqrt{2\left(2m^{2}+3\right)}+m}{m^{2}+2}\text{; }x=\frac{\sqrt{2}\left(m\sqrt{2m^{2}+3}+\sqrt{2}\right)}{m^{2}+2}\text{, }y=\frac{\sqrt{2\left(2m^{2}+3\right)}-m}{m^{2}+2}\text{, }&m\neq -\sqrt{2}i\text{ and }m\neq \sqrt{2}i\\x=\frac{5}{2}=2.5\text{, }y=\frac{3}{2m}\text{, }&m=-\sqrt{2}i\text{ or }m=\sqrt{2}i\end{matrix}\right.$$