Solve {l}{x^2+y^2=9}{x-3y=-1}

$$\left. \begin{array}{l}{ x ^ { 2 } + y ^ { 2 } = 9 }\\{ x - 3 y = - 1 }\end{array} \right.$$

Solve for x, y

$$x=\frac{-3\sqrt{89}-1}{10}\approx -2.93019434\text{, }y=\frac{3-\sqrt{89}}{10}\approx -0.643398113$$

$$x=\frac{3\sqrt{89}-1}{10}\approx 2.73019434\text{, }y=\frac{\sqrt{89}+3}{10}\approx 1.243398113$$

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.

$$x-3y=-1,y^{2}+x^{2}=9$$

Solve x-3y=-1 for x by isolating x on the left hand side of the equal sign.

$$x-3y=-1$$

Subtract -3y from both sides of the equation.

$$x=3y-1$$

Substitute 3y-1 for x in the other equation, y^{2}+x^{2}=9.

$$y^{2}+\left(3y-1\right)^{2}=9$$

Square 3y-1.

$$y^{2}+9y^{2}-6y+1=9$$

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

$$10y^{2}-6y+1=9$$

Subtract 9 from both sides of the equation.

$$10y^{2}-6y-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 3^{2} for a, 1\left(-1\right)\times 2\times 3 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 10\left(-8\right)}}{2\times 10}$$

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

$$y=\frac{-\left(-6\right)±\sqrt{36-4\times 10\left(-8\right)}}{2\times 10}$$

Multiply -4 times 1+1\times 3^{2}.

$$y=\frac{-\left(-6\right)±\sqrt{36-40\left(-8\right)}}{2\times 10}$$

Multiply -40 times -8.

$$y=\frac{-\left(-6\right)±\sqrt{36+320}}{2\times 10}$$

Add 36 to 320.

$$y=\frac{-\left(-6\right)±\sqrt{356}}{2\times 10}$$

Take the square root of 356.

$$y=\frac{-\left(-6\right)±2\sqrt{89}}{2\times 10}$$

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

$$y=\frac{6±2\sqrt{89}}{2\times 10}$$

Multiply 2 times 1+1\times 3^{2}.

$$y=\frac{6±2\sqrt{89}}{20}$$

Now solve the equation y=\frac{6±2\sqrt{89}}{20} when ± is plus. Add 6 to 2\sqrt{89}.

$$y=\frac{2\sqrt{89}+6}{20}$$

Divide 6+2\sqrt{89} by 20.

$$y=\frac{\sqrt{89}+3}{10}$$

Now solve the equation y=\frac{6±2\sqrt{89}}{20} when ± is minus. Subtract 2\sqrt{89} from 6.

$$y=\frac{6-2\sqrt{89}}{20}$$

Divide 6-2\sqrt{89} by 20.

$$y=\frac{3-\sqrt{89}}{10}$$

There are two solutions for y: \frac{3+\sqrt{89}}{10} and \frac{3-\sqrt{89}}{10}. Substitute \frac{3+\sqrt{89}}{10} for y in the equation x=3y-1 to find the corresponding solution for x that satisfies both equations.

$$x=3\times \frac{\sqrt{89}+3}{10}-1$$

Now substitute \frac{3-\sqrt{89}}{10} for y in the equation x=3y-1 and solve to find the corresponding solution for x that satisfies both equations.

$$x=3\times \frac{3-\sqrt{89}}{10}-1$$

The system is now solved.

$$x=3\times \frac{\sqrt{89}+3}{10}-1,y=\frac{\sqrt{89}+3}{10}\text{ or }x=3\times \frac{3-\sqrt{89}}{10}-1,y=\frac{3-\sqrt{89}}{10}$$