Solve {l}{x^2/4+y^2/4=6}{x+y=4}

$$\left\{ \begin{array} { l } { \frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 4 } = 6 } \\ { x + y = 4 } \end{array} \right.$$

Solve for x, y

$$x=2\sqrt{2}+2\approx 4.828427125\text{, }y=2-2\sqrt{2}\approx -0.828427125$$

$$x=2-2\sqrt{2}\approx -0.828427125\text{, }y=2\sqrt{2}+2\approx 4.828427125$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Multiply both sides of the equation by 4.

$$x^{2}+y^{2}=24$$

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.

$$x+y=4,y^{2}+x^{2}=24$$

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}=24.

$$y^{2}+\left(-y+4\right)^{2}=24$$

Square -y+4.

$$y^{2}+y^{2}-8y+16=24$$

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

$$2y^{2}-8y+16=24$$

Subtract 24 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}.

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

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

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

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

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

Multiply -8 times -8.

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

Add 64 to 64.

$$y=\frac{-\left(-8\right)±\sqrt{128}}{2\times 2}$$

Take the square root of 128.

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

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

$$y=\frac{8±8\sqrt{2}}{2\times 2}$$

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

$$y=\frac{8±8\sqrt{2}}{4}$$

Now solve the equation y=\frac{8±8\sqrt{2}}{4} when ± is plus. Add 8 to 8\sqrt{2}.

$$y=\frac{8\sqrt{2}+8}{4}$$

Divide 8+8\sqrt{2} by 4.

$$y=2\sqrt{2}+2$$

Now solve the equation y=\frac{8±8\sqrt{2}}{4} when ± is minus. Subtract 8\sqrt{2} from 8.

$$y=\frac{8-8\sqrt{2}}{4}$$

Divide 8-8\sqrt{2} by 4.

$$y=2-2\sqrt{2}$$

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

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

Now substitute 2-2\sqrt{2} for y in the equation x=-y+4 and solve to find the corresponding solution for x that satisfies both equations.

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

The system is now solved.

$$x=-\left(2\sqrt{2}+2\right)+4,y=2\sqrt{2}+2\text{ or }x=-\left(2-2\sqrt{2}\right)+4,y=2-2\sqrt{2}$$