Solve {l}{x^2+y^2=101}{2x+5y=1}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 101 } \\ { 2 x + 5 y = 1 } \end{array} \right.$$

Solve for x, y

$$x=\frac{20\sqrt{183}+2}{29}\approx 9.398447764\text{, }y=\frac{5-8\sqrt{183}}{29}\approx -3.559379106$$

$$x=\frac{2-20\sqrt{183}}{29}\approx -9.26051673\text{, }y=\frac{8\sqrt{183}+5}{29}\approx 3.904206692$$

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.

$$2x+5y=1,y^{2}+x^{2}=101$$

Solve 2x+5y=1 for x by isolating x on the left hand side of the equal sign.

$$2x+5y=1$$

Subtract 5y from both sides of the equation.

$$2x=-5y+1$$

Divide both sides by 2.

$$x=-\frac{5}{2}y+\frac{1}{2}$$

Substitute -\frac{5}{2}y+\frac{1}{2} for x in the other equation, y^{2}+x^{2}=101.

$$y^{2}+\left(-\frac{5}{2}y+\frac{1}{2}\right)^{2}=101$$

Square -\frac{5}{2}y+\frac{1}{2}.

$$y^{2}+\frac{25}{4}y^{2}-\frac{5}{2}y+\frac{1}{4}=101$$

Add y^{2} to \frac{25}{4}y^{2}.

$$\frac{29}{4}y^{2}-\frac{5}{2}y+\frac{1}{4}=101$$

Subtract 101 from both sides of the equation.

$$\frac{29}{4}y^{2}-\frac{5}{2}y-\frac{403}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-\frac{5}{2}\right)^{2} for a, 1\times \frac{1}{2}\left(-\frac{5}{2}\right)\times 2 for b, and -\frac{403}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-\frac{5}{2}\right)±\sqrt{\left(-\frac{5}{2}\right)^{2}-4\times \frac{29}{4}\left(-\frac{403}{4}\right)}}{2\times \frac{29}{4}}$$

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

$$y=\frac{-\left(-\frac{5}{2}\right)±\sqrt{\frac{25}{4}-4\times \frac{29}{4}\left(-\frac{403}{4}\right)}}{2\times \frac{29}{4}}$$

Multiply -4 times 1+1\left(-\frac{5}{2}\right)^{2}.

$$y=\frac{-\left(-\frac{5}{2}\right)±\sqrt{\frac{25}{4}-29\left(-\frac{403}{4}\right)}}{2\times \frac{29}{4}}$$

Multiply -29 times -\frac{403}{4}.

$$y=\frac{-\left(-\frac{5}{2}\right)±\sqrt{\frac{25+11687}{4}}}{2\times \frac{29}{4}}$$

Add \frac{25}{4} to \frac{11687}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{5}{2}\right)±\sqrt{2928}}{2\times \frac{29}{4}}$$

Take the square root of 2928.

$$y=\frac{-\left(-\frac{5}{2}\right)±4\sqrt{183}}{2\times \frac{29}{4}}$$

The opposite of 1\times \frac{1}{2}\left(-\frac{5}{2}\right)\times 2 is \frac{5}{2}.

$$y=\frac{\frac{5}{2}±4\sqrt{183}}{2\times \frac{29}{4}}$$

Multiply 2 times 1+1\left(-\frac{5}{2}\right)^{2}.

$$y=\frac{\frac{5}{2}±4\sqrt{183}}{\frac{29}{2}}$$

Now solve the equation y=\frac{\frac{5}{2}±4\sqrt{183}}{\frac{29}{2}} when ± is plus. Add \frac{5}{2} to 4\sqrt{183}.

$$y=\frac{4\sqrt{183}+\frac{5}{2}}{\frac{29}{2}}$$

Divide \frac{5}{2}+4\sqrt{183} by \frac{29}{2} by multiplying \frac{5}{2}+4\sqrt{183} by the reciprocal of \frac{29}{2}.

$$y=\frac{8\sqrt{183}+5}{29}$$

Now solve the equation y=\frac{\frac{5}{2}±4\sqrt{183}}{\frac{29}{2}} when ± is minus. Subtract 4\sqrt{183} from \frac{5}{2}.

$$y=\frac{\frac{5}{2}-4\sqrt{183}}{\frac{29}{2}}$$

Divide \frac{5}{2}-4\sqrt{183} by \frac{29}{2} by multiplying \frac{5}{2}-4\sqrt{183} by the reciprocal of \frac{29}{2}.

$$y=\frac{5-8\sqrt{183}}{29}$$

There are two solutions for y: \frac{5+8\sqrt{183}}{29} and \frac{5-8\sqrt{183}}{29}. Substitute \frac{5+8\sqrt{183}}{29} for y in the equation x=-\frac{5}{2}y+\frac{1}{2} to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{5}{2}\times \frac{8\sqrt{183}+5}{29}+\frac{1}{2}$$

Multiply -\frac{5}{2} times \frac{5+8\sqrt{183}}{29}.

$$x=\frac{-5\times \frac{8\sqrt{183}+5}{29}+1}{2}$$

Now substitute \frac{5-8\sqrt{183}}{29} for y in the equation x=-\frac{5}{2}y+\frac{1}{2} and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{5}{2}\times \frac{5-8\sqrt{183}}{29}+\frac{1}{2}$$

Multiply -\frac{5}{2} times \frac{5-8\sqrt{183}}{29}.

$$x=\frac{-5\times \frac{5-8\sqrt{183}}{29}+1}{2}$$

The system is now solved.

$$x=\frac{-5\times \frac{8\sqrt{183}+5}{29}+1}{2},y=\frac{8\sqrt{183}+5}{29}\text{ or }x=\frac{-5\times \frac{5-8\sqrt{183}}{29}+1}{2},y=\frac{5-8\sqrt{183}}{29}$$