Solve {l}{x+y=5-0}{x^2+y^2=18-(2)}

$$\left. \begin{array} { l } { x + y = 5 - 0 } \\ { x ^ { 2 } + y ^ { 2 } = 18 - ( 2 ) } \end{array} \right.$$

Solve for x, y

$$x=\frac{\sqrt{7}+5}{2}\approx 3.822875656\text{, }y=\frac{5-\sqrt{7}}{2}\approx 1.177124344$$

$$x=\frac{5-\sqrt{7}}{2}\approx 1.177124344\text{, }y=\frac{\sqrt{7}+5}{2}\approx 3.822875656$$

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.

$$x+y=5,y^{2}+x^{2}=16$$

Solve x+y=5 for x by isolating x on the left hand side of the equal sign.

$$x+y=5$$

Subtract y from both sides of the equation.

$$x=-y+5$$

Substitute -y+5 for x in the other equation, y^{2}+x^{2}=16.

$$y^{2}+\left(-y+5\right)^{2}=16$$

Square -y+5.

$$y^{2}+y^{2}-10y+25=16$$

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

$$2y^{2}-10y+25=16$$

Subtract 16 from both sides of the equation.

$$2y^{2}-10y+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\times 5\left(-1\right)\times 2 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 2\times 9}}{2\times 2}$$

Square 1\times 5\left(-1\right)\times 2.

$$y=\frac{-\left(-10\right)±\sqrt{100-4\times 2\times 9}}{2\times 2}$$

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

$$y=\frac{-\left(-10\right)±\sqrt{100-8\times 9}}{2\times 2}$$

Multiply -8 times 9.

$$y=\frac{-\left(-10\right)±\sqrt{100-72}}{2\times 2}$$

Add 100 to -72.

$$y=\frac{-\left(-10\right)±\sqrt{28}}{2\times 2}$$

Take the square root of 28.

$$y=\frac{-\left(-10\right)±2\sqrt{7}}{2\times 2}$$

The opposite of 1\times 5\left(-1\right)\times 2 is 10.

$$y=\frac{10±2\sqrt{7}}{2\times 2}$$

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

$$y=\frac{10±2\sqrt{7}}{4}$$

Now solve the equation y=\frac{10±2\sqrt{7}}{4} when ± is plus. Add 10 to 2\sqrt{7}.

$$y=\frac{2\sqrt{7}+10}{4}$$

Divide 10+2\sqrt{7} by 4.

$$y=\frac{\sqrt{7}+5}{2}$$

Now solve the equation y=\frac{10±2\sqrt{7}}{4} when ± is minus. Subtract 2\sqrt{7} from 10.

$$y=\frac{10-2\sqrt{7}}{4}$$

Divide 10-2\sqrt{7} by 4.

$$y=\frac{5-\sqrt{7}}{2}$$

There are two solutions for y: \frac{5+\sqrt{7}}{2} and \frac{5-\sqrt{7}}{2}. Substitute \frac{5+\sqrt{7}}{2} for y in the equation x=-y+5 to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{\sqrt{7}+5}{2}+5$$

Now substitute \frac{5-\sqrt{7}}{2} for y in the equation x=-y+5 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{5-\sqrt{7}}{2}+5$$

The system is now solved.

$$x=-\frac{\sqrt{7}+5}{2}+5,y=\frac{\sqrt{7}+5}{2}\text{ or }x=-\frac{5-\sqrt{7}}{2}+5,y=\frac{5-\sqrt{7}}{2}$$