Solve {c}{x^2+y^2=8}{x+y=4}

$$\left. \begin{array} { c } { x ^ { 2 } + y ^ { 2 } = 8 } \\ { x + y = 4 } \end{array} \right.$$

Solve for x, y

$$x=2$$

$$y=2$$

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.

$$x+y=4,y^{2}+x^{2}=8$$

Solve x+y=4 for x by isolating x on the left hand side of the equal sign.

$$x+y=4$$

Subtract y from both sides of the equation.

$$x=-y+4$$

Substitute -y+4 for x in the other equation, y^{2}+x^{2}=8.

$$y^{2}+\left(-y+4\right)^{2}=8$$

Square -y+4.

$$y^{2}+y^{2}-8y+16=8$$

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

$$2y^{2}-8y+16=8$$

Subtract 8 from both sides of the equation.

$$2y^{2}-8y+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\times 4\left(-1\right)\times 2 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 1\times 4\left(-1\right)\times 2.

$$y=\frac{-\left(-8\right)±\sqrt{64-4\times 2\times 8}}{2\times 2}$$

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

$$y=\frac{-\left(-8\right)±\sqrt{64-8\times 8}}{2\times 2}$$

Multiply -8 times 8.

$$y=\frac{-\left(-8\right)±\sqrt{64-64}}{2\times 2}$$

Add 64 to -64.

$$y=\frac{-\left(-8\right)±\sqrt{0}}{2\times 2}$$

Take the square root of 0.

$$y=-\frac{-8}{2\times 2}$$

The opposite of 1\times 4\left(-1\right)\times 2 is 8.

$$y=\frac{8}{2\times 2}$$

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

$$y=\frac{8}{4}$$

Divide 8 by 4.

$$y=2$$

There are two solutions for y: 2 and 2. Substitute 2 for y in the equation x=-y+4 to find the corresponding solution for x that satisfies both equations.

$$x=-2+4$$

Add -2 to 4.

$$x=2$$

The system is now solved.

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