Solve {l}{x^2+y^2=1}{x-5+y=0}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 1 } \\ { x - 5 + y = 0 } \end{array} \right.$$

Solve for x, y (complex solution)

$$x=\frac{5+\sqrt{23}i}{2}\approx 2.5+2.397915762i\text{, }y=\frac{-\sqrt{23}i+5}{2}\approx 2.5-2.397915762i$$

$$x=\frac{-\sqrt{23}i+5}{2}\approx 2.5-2.397915762i\text{, }y=\frac{5+\sqrt{23}i}{2}\approx 2.5+2.397915762i$$

Steps Using Substitution and Quadratic Formula

Solve x+y-5=0 for x by isolating x on the left hand side of the equal sign.

$$x+y-5=0$$

Add 5 to both sides of the equation.

$$x+y=5$$

Subtract y from both sides of the equation.

$$x=-y+5$$

Substitute -y+5 for x in the other equation, y^{2}+x^{2}=1.

$$y^{2}+\left(-y+5\right)^{2}=1$$

Square -y+5.

$$y^{2}+y^{2}-10y+25=1$$

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

$$2y^{2}-10y+25=1$$

Subtract 1 from both sides of the equation.

$$2y^{2}-10y+24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\times 5\left(-1\right)\times 2 for b, and 24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 2\times 24}}{2\times 2}$$

Square 1\times 5\left(-1\right)\times 2.

$$y=\frac{-\left(-10\right)±\sqrt{100-4\times 2\times 24}}{2\times 2}$$

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

$$y=\frac{-\left(-10\right)±\sqrt{100-8\times 24}}{2\times 2}$$

Multiply -8 times 24.

$$y=\frac{-\left(-10\right)±\sqrt{100-192}}{2\times 2}$$

Add 100 to -192.

$$y=\frac{-\left(-10\right)±\sqrt{-92}}{2\times 2}$$

Take the square root of -92.

$$y=\frac{-\left(-10\right)±2\sqrt{23}i}{2\times 2}$$

The opposite of 1\times 5\left(-1\right)\times 2 is 10.

$$y=\frac{10±2\sqrt{23}i}{2\times 2}$$

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

$$y=\frac{10±2\sqrt{23}i}{4}$$

Now solve the equation y=\frac{10±2\sqrt{23}i}{4} when ± is plus. Add 10 to 2i\sqrt{23}.

$$y=\frac{10+2\sqrt{23}i}{4}$$

Divide 10+2i\sqrt{23} by 4.

$$y=\frac{5+\sqrt{23}i}{2}$$

Now solve the equation y=\frac{10±2\sqrt{23}i}{4} when ± is minus. Subtract 2i\sqrt{23} from 10.

$$y=\frac{-2\sqrt{23}i+10}{4}$$

Divide 10-2i\sqrt{23} by 4.

$$y=\frac{-\sqrt{23}i+5}{2}$$

There are two solutions for y: \frac{5+i\sqrt{23}}{2} and \frac{5-i\sqrt{23}}{2}. Substitute \frac{5+i\sqrt{23}}{2} for y in the equation x=-y+5 to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{5+\sqrt{23}i}{2}+5$$

Now substitute \frac{5-i\sqrt{23}}{2} for y in the equation x=-y+5 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{-\sqrt{23}i+5}{2}+5$$

The system is now solved.

$$x=-\frac{5+\sqrt{23}i}{2}+5,y=\frac{5+\sqrt{23}i}{2}\text{ or }x=-\frac{-\sqrt{23}i+5}{2}+5,y=\frac{-\sqrt{23}i+5}{2}$$