Solve {l}{x^2+y^2=35}{x+2sqrt{2}=3sqrt{3}-y}

$$\left\{ \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 35 } \\ { x + 2 \sqrt { 2 } = 3 \sqrt { 3 } - y } \end{array} \right.$$

Solve for x, y

$$x=3\sqrt{3}\approx 5.196152423\text{, }y=-2\sqrt{2}\approx -2.828427125$$

$$x=-2\sqrt{2}\approx -2.828427125\text{, }y=3\sqrt{3}\approx 5.196152423$$

Steps Using Substitution and Quadratic Formula

Consider the second equation. Add y to both sides.

$$x+2\sqrt{2}+y=3\sqrt{3}$$

Subtract 2\sqrt{2} from both sides.

$$x+y=3\sqrt{3}-2\sqrt{2}$$

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.

$$x+y=3\sqrt{3}-2\sqrt{2},y^{2}+x^{2}=35$$

Solve x+y=3\sqrt{3}-2\sqrt{2} for x by isolating x on the left hand side of the equal sign.

$$x+y=3\sqrt{3}-2\sqrt{2}$$

Subtract y from both sides of the equation.

$$x=-y+3\sqrt{3}-2\sqrt{2}$$

Substitute -y+3\sqrt{3}-2\sqrt{2} for x in the other equation, y^{2}+x^{2}=35.

$$y^{2}+\left(-y+3\sqrt{3}-2\sqrt{2}\right)^{2}=35$$

Square -y+3\sqrt{3}-2\sqrt{2}.

$$y^{2}+y^{2}+\left(-2\left(3\sqrt{3}-2\sqrt{2}\right)\right)y+\left(3\sqrt{3}-2\sqrt{2}\right)^{2}=35$$

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

$$2y^{2}+\left(-2\left(3\sqrt{3}-2\sqrt{2}\right)\right)y+\left(3\sqrt{3}-2\sqrt{2}\right)^{2}=35$$

Subtract 35 from both sides of the equation.

$$2y^{2}+\left(-2\left(3\sqrt{3}-2\sqrt{2}\right)\right)y+\left(3\sqrt{3}-2\sqrt{2}\right)^{2}-35=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\left(-1\right)\times 2\left(3\sqrt{3}-2\sqrt{2}\right) for b, and -12\sqrt{6} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-2\left(3\sqrt{3}-2\sqrt{2}\right)\right)±\sqrt{\left(-2\left(3\sqrt{3}-2\sqrt{2}\right)\right)^{2}-4\times 2\left(-12\sqrt{6}\right)}}{2\times 2}$$

Square 1\left(-1\right)\times 2\left(3\sqrt{3}-2\sqrt{2}\right).

$$y=\frac{-\left(-2\left(3\sqrt{3}-2\sqrt{2}\right)\right)±\sqrt{140-48\sqrt{6}-4\times 2\left(-12\sqrt{6}\right)}}{2\times 2}$$

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

$$y=\frac{-\left(-2\left(3\sqrt{3}-2\sqrt{2}\right)\right)±\sqrt{140-48\sqrt{6}-8\left(-12\sqrt{6}\right)}}{2\times 2}$$

Multiply -8 times -12\sqrt{6}.

$$y=\frac{-\left(-2\left(3\sqrt{3}-2\sqrt{2}\right)\right)±\sqrt{140-48\sqrt{6}+96\sqrt{6}}}{2\times 2}$$

Add 140-48\sqrt{6} to 96\sqrt{6}.

$$y=\frac{-\left(-2\left(3\sqrt{3}-2\sqrt{2}\right)\right)±\sqrt{48\sqrt{6}+140}}{2\times 2}$$

Take the square root of 140+48\sqrt{6}.

$$y=\frac{-\left(-2\left(3\sqrt{3}-2\sqrt{2}\right)\right)±\left(4\sqrt{2}+6\sqrt{3}\right)}{2\times 2}$$

The opposite of 1\left(-1\right)\times 2\left(3\sqrt{3}-2\sqrt{2}\right) is 6\sqrt{3}-4\sqrt{2}.

$$y=\frac{6\sqrt{3}-4\sqrt{2}±\left(4\sqrt{2}+6\sqrt{3}\right)}{2\times 2}$$

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

$$y=\frac{6\sqrt{3}-4\sqrt{2}±\left(4\sqrt{2}+6\sqrt{3}\right)}{4}$$

Now solve the equation y=\frac{6\sqrt{3}-4\sqrt{2}±\left(4\sqrt{2}+6\sqrt{3}\right)}{4} when ± is plus. Add 6\sqrt{3}-4\sqrt{2} to 6\sqrt{3}+4\sqrt{2}.

$$y=\frac{12\sqrt{3}}{4}$$

Divide 12\sqrt{3} by 4.

$$y=3\sqrt{3}$$

Now solve the equation y=\frac{6\sqrt{3}-4\sqrt{2}±\left(4\sqrt{2}+6\sqrt{3}\right)}{4} when ± is minus. Subtract 6\sqrt{3}+4\sqrt{2} from 6\sqrt{3}-4\sqrt{2}.

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

Divide -8\sqrt{2} by 4.

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

There are two solutions for y: 3\sqrt{3} and -2\sqrt{2}. Substitute 3\sqrt{3} for y in the equation x=-y+3\sqrt{3}-2\sqrt{2} to find the corresponding solution for x that satisfies both equations.

$$x=-3\sqrt{3}+3\sqrt{3}-2\sqrt{2}$$

Add -3\sqrt{3} to 3\sqrt{3}-2\sqrt{2}.

$$x=3\sqrt{3}-2\sqrt{2}-3\sqrt{3}$$

Now substitute -2\sqrt{2} for y in the equation x=-y+3\sqrt{3}-2\sqrt{2} and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\left(-2\sqrt{2}\right)+3\sqrt{3}-2\sqrt{2}$$

Add -\left(-2\sqrt{2}\right) to 3\sqrt{3}-2\sqrt{2}.

$$x=3\sqrt{3}-2\sqrt{2}-\left(-2\sqrt{2}\right)$$

The system is now solved.

$$x=3\sqrt{3}-2\sqrt{2}-3\sqrt{3},y=3\sqrt{3}\text{ or }x=3\sqrt{3}-2\sqrt{2}-\left(-2\sqrt{2}\right),y=-2\sqrt{2}$$