Solve {l}{y=x+c}{x^2+y^2=13}

$$\left. \begin{array} { l } { y = x + c } \\ { x ^ { 2 } + y ^ { 2 } = 13 } \end{array} \right.$$

Solve for x, y (complex solution)

$$x=\frac{-\sqrt{26-c^{2}}-c}{2}\text{, }y=\frac{-\sqrt{26-c^{2}}+c}{2}$$

$$x=\frac{\sqrt{26-c^{2}}-c}{2}\text{, }y=\frac{\sqrt{26-c^{2}}+c}{2}$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Subtract x from both sides.

$$y-x=c$$

Subtract -x from both sides of the equation.

$$y=x+c$$

Substitute x+c for y in the other equation, x^{2}+y^{2}=13.

$$x^{2}+\left(x+c\right)^{2}=13$$

Square x+c.

$$x^{2}+x^{2}+2cx+c^{2}=13$$

Add x^{2} to x^{2}.

$$2x^{2}+2cx+c^{2}=13$$

Subtract 13 from both sides of the equation.

$$2x^{2}+2cx+c^{2}-13=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 1^{2} for a, 1\times 1\times 2c for b, and c^{2}-13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2c±\sqrt{\left(2c\right)^{2}-4\times 2\left(c^{2}-13\right)}}{2\times 2}$$

Square 1\times 1\times 2c.

$$x=\frac{-2c±\sqrt{4c^{2}-4\times 2\left(c^{2}-13\right)}}{2\times 2}$$

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

$$x=\frac{-2c±\sqrt{4c^{2}-8\left(c^{2}-13\right)}}{2\times 2}$$

Multiply -8 times c^{2}-13.

$$x=\frac{-2c±\sqrt{4c^{2}+104-8c^{2}}}{2\times 2}$$

Add 4c^{2} to -8c^{2}+104.

$$x=\frac{-2c±\sqrt{104-4c^{2}}}{2\times 2}$$

Take the square root of 104-4c^{2}.

$$x=\frac{-2c±2\sqrt{26-c^{2}}}{2\times 2}$$

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

$$x=\frac{-2c±2\sqrt{26-c^{2}}}{4}$$

Now solve the equation x=\frac{-2c±2\sqrt{26-c^{2}}}{4} when ± is plus. Add -2c to 2\sqrt{26-c^{2}}.

$$x=\frac{2\sqrt{26-c^{2}}-2c}{4}$$

Divide -2c+2\sqrt{26-c^{2}} by 4.

$$x=\frac{\sqrt{26-c^{2}}-c}{2}$$

Now solve the equation x=\frac{-2c±2\sqrt{26-c^{2}}}{4} when ± is minus. Subtract 2\sqrt{26-c^{2}} from -2c.

$$x=\frac{-2\sqrt{26-c^{2}}-2c}{4}$$

Divide -2c-2\sqrt{26-c^{2}} by 4.

$$x=\frac{-\sqrt{26-c^{2}}-c}{2}$$

There are two solutions for x: \frac{-c+\sqrt{26-c^{2}}}{2} and \frac{-c-\sqrt{26-c^{2}}}{2}. Substitute \frac{-c+\sqrt{26-c^{2}}}{2} for x in the equation y=x+c to find the corresponding solution for y that satisfies both equations.

$$y=\frac{\sqrt{26-c^{2}}-c}{2}+c$$

Now substitute \frac{-c-\sqrt{26-c^{2}}}{2} for x in the equation y=x+c and solve to find the corresponding solution for y that satisfies both equations.

$$y=\frac{-\sqrt{26-c^{2}}-c}{2}+c$$

The system is now solved.

$$y=\frac{\sqrt{26-c^{2}}-c}{2}+c,x=\frac{\sqrt{26-c^{2}}-c}{2}\text{ or }y=\frac{-\sqrt{26-c^{2}}-c}{2}+c,x=\frac{-\sqrt{26-c^{2}}-c}{2}$$

Solve for x, y

$$x=\frac{-\sqrt{26-c^{2}}-c}{2}\text{, }y=\frac{-\sqrt{26-c^{2}}+c}{2}$$

$$x=\frac{\sqrt{26-c^{2}}-c}{2}\text{, }y=\frac{\sqrt{26-c^{2}}+c}{2}\text{, }|c|\leq \sqrt{26}$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Subtract x from both sides.

$$y-x=c$$

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.

$$y-x=c,x^{2}+y^{2}=13$$

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

$$y-x=c$$

Subtract -x from both sides of the equation.

$$y=x+c$$

Substitute x+c for y in the other equation, x^{2}+y^{2}=13.

$$x^{2}+\left(x+c\right)^{2}=13$$

Square x+c.

$$x^{2}+x^{2}+2cx+c^{2}=13$$

Add x^{2} to x^{2}.

$$2x^{2}+2cx+c^{2}=13$$

Subtract 13 from both sides of the equation.

$$2x^{2}+2cx+c^{2}-13=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 1^{2} for a, 1\times 1\times 2c for b, and c^{2}-13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2c±\sqrt{\left(2c\right)^{2}-4\times 2\left(c^{2}-13\right)}}{2\times 2}$$

Square 1\times 1\times 2c.

$$x=\frac{-2c±\sqrt{4c^{2}-4\times 2\left(c^{2}-13\right)}}{2\times 2}$$

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

$$x=\frac{-2c±\sqrt{4c^{2}-8\left(c^{2}-13\right)}}{2\times 2}$$

Multiply -8 times c^{2}-13.

$$x=\frac{-2c±\sqrt{4c^{2}+104-8c^{2}}}{2\times 2}$$

Add 4c^{2} to -8c^{2}+104.

$$x=\frac{-2c±\sqrt{104-4c^{2}}}{2\times 2}$$

Take the square root of 104-4c^{2}.

$$x=\frac{-2c±2\sqrt{26-c^{2}}}{2\times 2}$$

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

$$x=\frac{-2c±2\sqrt{26-c^{2}}}{4}$$

Now solve the equation x=\frac{-2c±2\sqrt{26-c^{2}}}{4} when ± is plus. Add -2c to 2\sqrt{26-c^{2}}.

$$x=\frac{2\sqrt{26-c^{2}}-2c}{4}$$

Divide -2c+2\sqrt{26-c^{2}} by 4.

$$x=\frac{\sqrt{26-c^{2}}-c}{2}$$

Now solve the equation x=\frac{-2c±2\sqrt{26-c^{2}}}{4} when ± is minus. Subtract 2\sqrt{26-c^{2}} from -2c.

$$x=\frac{-2\sqrt{26-c^{2}}-2c}{4}$$

Divide -2c-2\sqrt{26-c^{2}} by 4.

$$x=\frac{-\sqrt{26-c^{2}}-c}{2}$$

There are two solutions for x: \frac{-c+\sqrt{26-c^{2}}}{2} and \frac{-c-\sqrt{26-c^{2}}}{2}. Substitute \frac{-c+\sqrt{26-c^{2}}}{2} for x in the equation y=x+c to find the corresponding solution for y that satisfies both equations.

$$y=\frac{\sqrt{26-c^{2}}-c}{2}+c$$

Now substitute \frac{-c-\sqrt{26-c^{2}}}{2} for x in the equation y=x+c and solve to find the corresponding solution for y that satisfies both equations.

$$y=\frac{-\sqrt{26-c^{2}}-c}{2}+c$$

The system is now solved.

$$y=\frac{\sqrt{26-c^{2}}-c}{2}+c,x=\frac{\sqrt{26-c^{2}}-c}{2}\text{ or }y=\frac{-\sqrt{26-c^{2}}-c}{2}+c,x=\frac{-\sqrt{26-c^{2}}-c}{2}$$