Solve {l}{x^2+y^2=128}{x+3y=19}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 128 } \\ { x + 3 y = 19 } \end{array} \right.$$

Solve for x, y

$$x=\frac{3\sqrt{919}+19}{10}\approx 10.994503835\text{, }y=\frac{57-\sqrt{919}}{10}\approx 2.668498722$$

$$x=\frac{19-3\sqrt{919}}{10}\approx -7.194503835\text{, }y=\frac{\sqrt{919}+57}{10}\approx 8.731501278$$

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.

$$x+3y=19,y^{2}+x^{2}=128$$

Solve x+3y=19 for x by isolating x on the left hand side of the equal sign.

$$x+3y=19$$

Subtract 3y from both sides of the equation.

$$x=-3y+19$$

Substitute -3y+19 for x in the other equation, y^{2}+x^{2}=128.

$$y^{2}+\left(-3y+19\right)^{2}=128$$

Square -3y+19.

$$y^{2}+9y^{2}-114y+361=128$$

Add y^{2} to 9y^{2}.

$$10y^{2}-114y+361=128$$

Subtract 128 from both sides of the equation.

$$10y^{2}-114y+233=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-3\right)^{2} for a, 1\times 19\left(-3\right)\times 2 for b, and 233 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-114\right)±\sqrt{\left(-114\right)^{2}-4\times 10\times 233}}{2\times 10}$$

Square 1\times 19\left(-3\right)\times 2.

$$y=\frac{-\left(-114\right)±\sqrt{12996-4\times 10\times 233}}{2\times 10}$$

Multiply -4 times 1+1\left(-3\right)^{2}.

$$y=\frac{-\left(-114\right)±\sqrt{12996-40\times 233}}{2\times 10}$$

Multiply -40 times 233.

$$y=\frac{-\left(-114\right)±\sqrt{12996-9320}}{2\times 10}$$

Add 12996 to -9320.

$$y=\frac{-\left(-114\right)±\sqrt{3676}}{2\times 10}$$

Take the square root of 3676.

$$y=\frac{-\left(-114\right)±2\sqrt{919}}{2\times 10}$$

The opposite of 1\times 19\left(-3\right)\times 2 is 114.

$$y=\frac{114±2\sqrt{919}}{2\times 10}$$

Multiply 2 times 1+1\left(-3\right)^{2}.

$$y=\frac{114±2\sqrt{919}}{20}$$

Now solve the equation y=\frac{114±2\sqrt{919}}{20} when ± is plus. Add 114 to 2\sqrt{919}.

$$y=\frac{2\sqrt{919}+114}{20}$$

Divide 114+2\sqrt{919} by 20.

$$y=\frac{\sqrt{919}+57}{10}$$

Now solve the equation y=\frac{114±2\sqrt{919}}{20} when ± is minus. Subtract 2\sqrt{919} from 114.

$$y=\frac{114-2\sqrt{919}}{20}$$

Divide 114-2\sqrt{919} by 20.

$$y=\frac{57-\sqrt{919}}{10}$$

There are two solutions for y: \frac{57+\sqrt{919}}{10} and \frac{57-\sqrt{919}}{10}. Substitute \frac{57+\sqrt{919}}{10} for y in the equation x=-3y+19 to find the corresponding solution for x that satisfies both equations.

$$x=-3\times \frac{\sqrt{919}+57}{10}+19$$

Now substitute \frac{57-\sqrt{919}}{10} for y in the equation x=-3y+19 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-3\times \frac{57-\sqrt{919}}{10}+19$$

The system is now solved.

$$x=-3\times \frac{\sqrt{919}+57}{10}+19,y=\frac{\sqrt{919}+57}{10}\text{ or }x=-3\times \frac{57-\sqrt{919}}{10}+19,y=\frac{57-\sqrt{919}}{10}$$