Solve {l}{ax+y=1}{x^2+y^2=1}

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

Solve for x, y

$$x=0\text{, }y=1$$

$$x=\frac{2a}{a^{2}+1}\text{, }y=\frac{1-a^{2}}{a^{2}+1}$$

Steps Using Substitution and Quadratic Formula

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.

$$ax+y=1,y^{2}+x^{2}=1$$

Solve ax+y=1 for x by isolating x on the left hand side of the equal sign.

$$ax+y=1$$

Subtract y from both sides of the equation.

$$ax=-y+1$$

Divide both sides by a.

$$x=\left(-\frac{1}{a}\right)y+\frac{1}{a}$$

Substitute \left(-\frac{1}{a}\right)y+\frac{1}{a} for x in the other equation, y^{2}+x^{2}=1.

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

Square \left(-\frac{1}{a}\right)y+\frac{1}{a}.

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

Add y^{2} to \left(-\frac{1}{a}\right)^{2}y^{2}.

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

Subtract 1 from both sides of the equation.

$$\left(1+\left(-\frac{1}{a}\right)^{2}\right)y^{2}+2\left(-\frac{1}{a}\right)\times \frac{1}{a}y+\left(\frac{1}{a}\right)^{2}-1=0$$

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

$$y=\frac{-2\left(-\frac{1}{a}\right)\times \frac{1}{a}±\sqrt{\left(2\left(-\frac{1}{a}\right)\times \frac{1}{a}\right)^{2}-4\left(1+\left(-\frac{1}{a}\right)^{2}\right)\left(-1+\frac{1}{a^{2}}\right)}}{2\left(1+\left(-\frac{1}{a}\right)^{2}\right)}$$

Square 1\times 2\left(-\frac{1}{a}\right)\times \frac{1}{a}.

$$y=\frac{-2\left(-\frac{1}{a}\right)\times \frac{1}{a}±\sqrt{\frac{4}{a^{4}}-4\left(1+\left(-\frac{1}{a}\right)^{2}\right)\left(-1+\frac{1}{a^{2}}\right)}}{2\left(1+\left(-\frac{1}{a}\right)^{2}\right)}$$

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

$$y=\frac{-2\left(-\frac{1}{a}\right)\times \frac{1}{a}±\sqrt{\frac{4}{a^{4}}+\left(-4-\frac{4}{a^{2}}\right)\left(-1+\frac{1}{a^{2}}\right)}}{2\left(1+\left(-\frac{1}{a}\right)^{2}\right)}$$

Multiply -4-\frac{4}{a^{2}} times -1+\frac{1}{a^{2}}.

$$y=\frac{-2\left(-\frac{1}{a}\right)\times \frac{1}{a}±\sqrt{\frac{4}{a^{4}}+4-\frac{4}{a^{4}}}}{2\left(1+\left(-\frac{1}{a}\right)^{2}\right)}$$

Add \frac{4}{a^{4}} to -\frac{4}{a^{4}}+4.

$$y=\frac{-2\left(-\frac{1}{a}\right)\times \frac{1}{a}±\sqrt{4}}{2\left(1+\left(-\frac{1}{a}\right)^{2}\right)}$$

Take the square root of 4.

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

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

$$y=\frac{\frac{2}{a^{2}}±2}{2+\frac{2}{a^{2}}}$$

Now solve the equation y=\frac{\frac{2}{a^{2}}±2}{2+\frac{2}{a^{2}}} when ± is plus. Add \frac{2}{a^{2}} to 2.

$$y=\frac{2+\frac{2}{a^{2}}}{2+\frac{2}{a^{2}}}$$

Divide 2+\frac{2}{a^{2}} by 2+\frac{2}{a^{2}}.

$$y=1$$

Now solve the equation y=\frac{\frac{2}{a^{2}}±2}{2+\frac{2}{a^{2}}} when ± is minus. Subtract 2 from \frac{2}{a^{2}}.

$$y=\frac{-2+\frac{2}{a^{2}}}{2+\frac{2}{a^{2}}}$$

Divide \frac{2}{a^{2}}-2 by 2+\frac{2}{a^{2}}.

$$y=-\frac{a^{2}-1}{a^{2}+1}$$

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

$$x=\frac{-1+1}{a}$$

Add 1\left(-\frac{1}{a}\right) to \frac{1}{a}.

$$x=\frac{1-1}{a}$$

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

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

Multiply -\frac{1}{a} times -\frac{a^{2}-1}{1+a^{2}}.

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

The system is now solved.

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

Solve for x, y (complex solution)

$$\left\{\begin{matrix}\\x=0\text{, }y=1\text{, }&\text{unconditionally}\\x=\frac{2a}{a^{2}+1}\text{, }y=\frac{1-a^{2}}{a^{2}+1}\text{, }&a\neq -i\text{ and }a\neq i\end{matrix}\right.$$