Solve {c}{x^2+y^2=4}{y-x=1}

$$\left. \begin{array} { c } { x ^ { 2 } + y ^ { 2 } = 4 } \\ { y - x = 1 } \end{array} \right.$$

Solve for x, y

$$x=\frac{-\sqrt{7}-1}{2}\approx -1.822875656\text{, }y=\frac{1-\sqrt{7}}{2}\approx -0.822875656$$

$$x=\frac{\sqrt{7}-1}{2}\approx 0.822875656\text{, }y=\frac{\sqrt{7}+1}{2}\approx 1.822875656$$

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.

$$y-x=1,x^{2}+y^{2}=4$$

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

$$y-x=1$$

Subtract -x from both sides of the equation.

$$y=x+1$$

Substitute x+1 for y in the other equation, x^{2}+y^{2}=4.

$$x^{2}+\left(x+1\right)^{2}=4$$

Square x+1.

$$x^{2}+x^{2}+2x+1=4$$

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

$$2x^{2}+2x+1=4$$

Subtract 4 from both sides of the equation.

$$2x^{2}+2x-3=0$$

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

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

Square 1\times 1\times 1\times 2.

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

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

$$x=\frac{-2±\sqrt{4-8\left(-3\right)}}{2\times 2}$$

Multiply -8 times -3.

$$x=\frac{-2±\sqrt{4+24}}{2\times 2}$$

Add 4 to 24.

$$x=\frac{-2±\sqrt{28}}{2\times 2}$$

Take the square root of 28.

$$x=\frac{-2±2\sqrt{7}}{2\times 2}$$

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

$$x=\frac{-2±2\sqrt{7}}{4}$$

Now solve the equation x=\frac{-2±2\sqrt{7}}{4} when ± is plus. Add -2 to 2\sqrt{7}.

$$x=\frac{2\sqrt{7}-2}{4}$$

Divide -2+2\sqrt{7} by 4.

$$x=\frac{\sqrt{7}-1}{2}$$

Now solve the equation x=\frac{-2±2\sqrt{7}}{4} when ± is minus. Subtract 2\sqrt{7} from -2.

$$x=\frac{-2\sqrt{7}-2}{4}$$

Divide -2-2\sqrt{7} by 4.

$$x=\frac{-\sqrt{7}-1}{2}$$

There are two solutions for x: \frac{-1+\sqrt{7}}{2} and \frac{-1-\sqrt{7}}{2}. Substitute \frac{-1+\sqrt{7}}{2} for x in the equation y=x+1 to find the corresponding solution for y that satisfies both equations.

$$y=\frac{\sqrt{7}-1}{2}+1$$

Now substitute \frac{-1-\sqrt{7}}{2} for x in the equation y=x+1 and solve to find the corresponding solution for y that satisfies both equations.

$$y=\frac{-\sqrt{7}-1}{2}+1$$

The system is now solved.

$$y=\frac{\sqrt{7}-1}{2}+1,x=\frac{\sqrt{7}-1}{2}\text{ or }y=\frac{-\sqrt{7}-1}{2}+1,x=\frac{-\sqrt{7}-1}{2}$$