Solve {l}{y=-mx+b}{x^2/2+y^2=1}

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

Solve for x, y

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

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

Steps Using Substitution and Quadratic Formula

Consider the first equation. Subtract \left(-m\right)x from both sides.

$$y-\left(-m\right)x=b$$

Multiply -1 and -1 to get 1.

$$y+mx=b$$

Consider the second equation. Multiply both sides of the equation by 2.

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

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.

$$y+mx=b,x^{2}+2y^{2}=2$$

Solve y+mx=b for y by isolating y on the left hand side of the equal sign.

$$y+mx=b$$

Subtract mx from both sides of the equation.

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

Substitute \left(-m\right)x+b for y in the other equation, x^{2}+2y^{2}=2.

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

Square \left(-m\right)x+b.

$$x^{2}+2\left(\left(-m\right)^{2}x^{2}+2\left(-m\right)bx+b^{2}\right)=2$$

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

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

Add x^{2} to 2\left(-m\right)^{2}x^{2}.

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

Subtract 2 from both sides of the equation.

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

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

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

Square 2\times 2\left(-m\right)b.

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

Multiply -4 times 1+2\left(-m\right)^{2}.

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

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

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

Add 16m^{2}b^{2} to -8\left(1+2m^{2}\right)\left(b^{2}-1\right).

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

Take the square root of 16m^{2}-8b^{2}+8.

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

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

$$x=\frac{4bm±2\sqrt{4m^{2}-2b^{2}+2}}{4m^{2}+2}$$

Now solve the equation x=\frac{4bm±2\sqrt{4m^{2}-2b^{2}+2}}{4m^{2}+2} when ± is plus. Add 4mb to 2\sqrt{4m^{2}-2b^{2}+2}.

$$x=\frac{4bm+2\sqrt{4m^{2}-2b^{2}+2}}{4m^{2}+2}$$

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

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

Now solve the equation x=\frac{4bm±2\sqrt{4m^{2}-2b^{2}+2}}{4m^{2}+2} when ± is minus. Subtract 2\sqrt{4m^{2}-2b^{2}+2} from 4mb.

$$x=\frac{4bm-2\sqrt{4m^{2}-2b^{2}+2}}{4m^{2}+2}$$

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

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

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

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

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

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

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

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

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

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

The system is now solved.

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

Solve for x, y (complex solution)

$$\left\{\begin{matrix}x=\frac{\sqrt{2}\left(\sqrt{2}bm-\sqrt{2m^{2}-b^{2}+1}\right)}{2m^{2}+1}\text{, }y=\frac{m\sqrt{2\left(2m^{2}-b^{2}+1\right)}+b}{2m^{2}+1}\text{; }x=\frac{\sqrt{2}\left(\sqrt{2}bm+\sqrt{2m^{2}-b^{2}+1}\right)}{2m^{2}+1}\text{, }y=\frac{-m\sqrt{2\left(2m^{2}-b^{2}+1\right)}+b}{2m^{2}+1}\text{, }&m\neq -\frac{\sqrt{2}i}{2}\text{ and }m\neq \frac{\sqrt{2}i}{2}\\x=\frac{b^{2}-1}{2bm}\text{, }y=\frac{b^{2}+1}{2b}\text{, }&b\neq 0\text{ and }\left(m=-\frac{\sqrt{2}i}{2}\text{ or }m=\frac{\sqrt{2}i}{2}\right)\end{matrix}\right.$$