Solve {l}{x+y=2}{x^2+y^2=34}

$$\left\{ \begin{array} { l } { x + y = 2 } \\ { x ^ { 2 } + y ^ { 2 } = 34 } \end{array} \right.$$

Solve for x, y

$$x=5\text{, }y=-3$$

$$x=-3\text{, }y=5$$

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.

$$x+y=2,y^{2}+x^{2}=34$$

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

$$x+y=2$$

Subtract y from both sides of the equation.

$$x=-y+2$$

Substitute -y+2 for x in the other equation, y^{2}+x^{2}=34.

$$y^{2}+\left(-y+2\right)^{2}=34$$

Square -y+2.

$$y^{2}+y^{2}-4y+4=34$$

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

$$2y^{2}-4y+4=34$$

Subtract 34 from both sides of the equation.

$$2y^{2}-4y-30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\times 2\left(-1\right)\times 2 for b, and -30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 2\left(-30\right)}}{2\times 2}$$

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

$$y=\frac{-\left(-4\right)±\sqrt{16-4\times 2\left(-30\right)}}{2\times 2}$$

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

$$y=\frac{-\left(-4\right)±\sqrt{16-8\left(-30\right)}}{2\times 2}$$

Multiply -8 times -30.

$$y=\frac{-\left(-4\right)±\sqrt{16+240}}{2\times 2}$$

Add 16 to 240.

$$y=\frac{-\left(-4\right)±\sqrt{256}}{2\times 2}$$

Take the square root of 256.

$$y=\frac{-\left(-4\right)±16}{2\times 2}$$

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

$$y=\frac{4±16}{2\times 2}$$

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

$$y=\frac{4±16}{4}$$

Now solve the equation y=\frac{4±16}{4} when ± is plus. Add 4 to 16.

$$y=\frac{20}{4}$$

Divide 20 by 4.

$$y=5$$

Now solve the equation y=\frac{4±16}{4} when ± is minus. Subtract 16 from 4.

$$y=-\frac{12}{4}$$

Divide -12 by 4.

$$y=-3$$

There are two solutions for y: 5 and -3. Substitute 5 for y in the equation x=-y+2 to find the corresponding solution for x that satisfies both equations.

$$x=-5+2$$

Add -5 to 2.

$$x=-3$$

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

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

Multiply -1 times -3.

$$x=3+2$$

Add -3\left(-1\right) to 2.

$$x=5$$

The system is now solved.

$$x=-3,y=5\text{ or }x=5,y=-3$$