Solve {l}{y=x+2}{x^2+y^2-4=0}

$$\left\{ \begin{array} { l } { y = x + 2 } \\ { x ^ { 2 } + y ^ { 2 } - 4 = 0 } \end{array} \right.$$

Solve for y, x

$$x=0\text{, }y=2$$

$$x=-2\text{, }y=0$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Subtract x from both sides.

$$y-x=2$$

Consider the second equation. Add 4 to both sides. Anything plus zero gives itself.

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

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.

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

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

$$y-x=2$$

Subtract -x from both sides of the equation.

$$y=x+2$$

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

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

Square x+2.

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

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

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

Subtract 4 from both sides of the equation.

$$2x^{2}+4x=0$$

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

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

Take the square root of 4^{2}.

$$x=\frac{-4±4}{2\times 2}$$

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

$$x=\frac{-4±4}{4}$$

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

$$x=\frac{0}{4}$$

Divide 0 by 4.

$$x=0$$

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

$$x=-\frac{8}{4}$$

Divide -8 by 4.

$$x=-2$$

There are two solutions for x: 0 and -2. Substitute 0 for x in the equation y=x+2 to find the corresponding solution for y that satisfies both equations.

$$y=2$$

Now substitute -2 for x in the equation y=x+2 and solve to find the corresponding solution for y that satisfies both equations.

$$y=-2+2$$

Add -2 to 2.

$$y=0$$

The system is now solved.

$$y=2,x=0\text{ or }y=0,x=-2$$