Solve {r}{x^2+y^2=1}{2x-y=3}

$$\left.\begin{array} { r } { x ^ { 2 } + y ^ { 2 } = 1 } \\ { 2 x - y = 3 } \end{array} \right\}$$

Solve for x, y (complex solution)

$$x=\frac{6}{5}-\frac{2}{5}i=1.2-0.4i\text{, }y=-\frac{3}{5}-\frac{4}{5}i=-0.6-0.8i$$

$$x=\frac{6}{5}+\frac{2}{5}i=1.2+0.4i\text{, }y=-\frac{3}{5}+\frac{4}{5}i=-0.6+0.8i$$

Steps Using Substitution and Quadratic Formula

Solve 2x-y=3 for x by isolating x on the left hand side of the equal sign.

$$2x-y=3$$

Subtract -y from both sides of the equation.

$$2x=y+3$$

Divide both sides by 2.

$$x=\frac{1}{2}y+\frac{3}{2}$$

Substitute \frac{1}{2}y+\frac{3}{2} for x in the other equation, y^{2}+x^{2}=1.

$$y^{2}+\left(\frac{1}{2}y+\frac{3}{2}\right)^{2}=1$$

Square \frac{1}{2}y+\frac{3}{2}.

$$y^{2}+\frac{1}{4}y^{2}+\frac{3}{2}y+\frac{9}{4}=1$$

Add y^{2} to \frac{1}{4}y^{2}.

$$\frac{5}{4}y^{2}+\frac{3}{2}y+\frac{9}{4}=1$$

Subtract 1 from both sides of the equation.

$$\frac{5}{4}y^{2}+\frac{3}{2}y+\frac{5}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times \left(\frac{1}{2}\right)^{2} for a, 1\times \frac{3}{2}\times \frac{1}{2}\times 2 for b, and \frac{5}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\frac{3}{2}±\sqrt{\left(\frac{3}{2}\right)^{2}-4\times \frac{5}{4}\times \frac{5}{4}}}{2\times \frac{5}{4}}$$

Square 1\times \frac{3}{2}\times \frac{1}{2}\times 2.

$$y=\frac{-\frac{3}{2}±\sqrt{\frac{9}{4}-4\times \frac{5}{4}\times \frac{5}{4}}}{2\times \frac{5}{4}}$$

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

$$y=\frac{-\frac{3}{2}±\sqrt{\frac{9}{4}-5\times \frac{5}{4}}}{2\times \frac{5}{4}}$$

Multiply -5 times \frac{5}{4}.

$$y=\frac{-\frac{3}{2}±\sqrt{\frac{9-25}{4}}}{2\times \frac{5}{4}}$$

Add \frac{9}{4} to -\frac{25}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\frac{3}{2}±\sqrt{-4}}{2\times \frac{5}{4}}$$

Take the square root of -4.

$$y=\frac{-\frac{3}{2}±2i}{2\times \frac{5}{4}}$$

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

$$y=\frac{-\frac{3}{2}±2i}{\frac{5}{2}}$$

Now solve the equation y=\frac{-\frac{3}{2}±2i}{\frac{5}{2}} when ± is plus. Add -\frac{3}{2} to 2i.

$$y=\frac{-\frac{3}{2}+2i}{\frac{5}{2}}$$

Divide -\frac{3}{2}+2i by \frac{5}{2} by multiplying -\frac{3}{2}+2i by the reciprocal of \frac{5}{2}.

$$y=-\frac{3}{5}+\frac{4}{5}i$$

Now solve the equation y=\frac{-\frac{3}{2}±2i}{\frac{5}{2}} when ± is minus. Subtract 2i from -\frac{3}{2}.

$$y=\frac{-\frac{3}{2}-2i}{\frac{5}{2}}$$

Divide -\frac{3}{2}-2i by \frac{5}{2} by multiplying -\frac{3}{2}-2i by the reciprocal of \frac{5}{2}.

$$y=-\frac{3}{5}-\frac{4}{5}i$$

There are two solutions for y: -\frac{3}{5}+\frac{4}{5}i and -\frac{3}{5}-\frac{4}{5}i. Substitute -\frac{3}{5}+\frac{4}{5}i for y in the equation x=\frac{1}{2}y+\frac{3}{2} to find the corresponding solution for x that satisfies both equations.

$$x=\frac{1}{2}\left(-\frac{3}{5}+\frac{4}{5}i\right)+\frac{3}{2}$$

Multiply \frac{1}{2} times -\frac{3}{5}+\frac{4}{5}i.

$$x=-\frac{3}{10}+\frac{2}{5}i+\frac{3}{2}$$

Add \left(-\frac{3}{5}+\frac{4}{5}i\right)\times \frac{1}{2} to \frac{3}{2}.

$$x=\frac{6}{5}+\frac{2}{5}i$$

Now substitute -\frac{3}{5}-\frac{4}{5}i for y in the equation x=\frac{1}{2}y+\frac{3}{2} and solve to find the corresponding solution for x that satisfies both equations.

$$x=\frac{1}{2}\left(-\frac{3}{5}-\frac{4}{5}i\right)+\frac{3}{2}$$

Multiply \frac{1}{2} times -\frac{3}{5}-\frac{4}{5}i.

$$x=-\frac{3}{10}-\frac{2}{5}i+\frac{3}{2}$$

Add \left(-\frac{3}{5}-\frac{4}{5}i\right)\times \frac{1}{2} to \frac{3}{2}.

$$x=\frac{6}{5}-\frac{2}{5}i$$

The system is now solved.

$$x=\frac{6}{5}+\frac{2}{5}i,y=-\frac{3}{5}+\frac{4}{5}i\text{ or }x=\frac{6}{5}-\frac{2}{5}i,y=-\frac{3}{5}-\frac{4}{5}i$$