Solve {l}{x^2+y^2=10}{2x+y=6}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 10 } \\ { 2 x + y = 6 } \end{array} \right.$$

Solve for x, y

$$x=\frac{\sqrt{14}+12}{5}\approx 3.148331477\text{, }y=\frac{6-2\sqrt{14}}{5}\approx -0.296662955$$

$$x=\frac{12-\sqrt{14}}{5}\approx 1.651668523\text{, }y=\frac{2\sqrt{14}+6}{5}\approx 2.696662955$$

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.

$$2x+y=6,y^{2}+x^{2}=10$$

Solve 2x+y=6 for x by isolating x on the left hand side of the equal sign.

$$2x+y=6$$

Subtract y from both sides of the equation.

$$2x=-y+6$$

Divide both sides by 2.

$$x=-\frac{1}{2}y+3$$

Substitute -\frac{1}{2}y+3 for x in the other equation, y^{2}+x^{2}=10.

$$y^{2}+\left(-\frac{1}{2}y+3\right)^{2}=10$$

Square -\frac{1}{2}y+3.

$$y^{2}+\frac{1}{4}y^{2}-3y+9=10$$

Add y^{2} to \frac{1}{4}y^{2}.

$$\frac{5}{4}y^{2}-3y+9=10$$

Subtract 10 from both sides of the equation.

$$\frac{5}{4}y^{2}-3y-1=0$$

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

$$y=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times \frac{5}{4}\left(-1\right)}}{2\times \frac{5}{4}}$$

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

$$y=\frac{-\left(-3\right)±\sqrt{9-4\times \frac{5}{4}\left(-1\right)}}{2\times \frac{5}{4}}$$

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

$$y=\frac{-\left(-3\right)±\sqrt{9-5\left(-1\right)}}{2\times \frac{5}{4}}$$

Multiply -5 times -1.

$$y=\frac{-\left(-3\right)±\sqrt{9+5}}{2\times \frac{5}{4}}$$

Add 9 to 5.

$$y=\frac{-\left(-3\right)±\sqrt{14}}{2\times \frac{5}{4}}$$

The opposite of 1\times 3\left(-\frac{1}{2}\right)\times 2 is 3.

$$y=\frac{3±\sqrt{14}}{2\times \frac{5}{4}}$$

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

$$y=\frac{3±\sqrt{14}}{\frac{5}{2}}$$

Now solve the equation y=\frac{3±\sqrt{14}}{\frac{5}{2}} when ± is plus. Add 3 to \sqrt{14}.

$$y=\frac{\sqrt{14}+3}{\frac{5}{2}}$$

Divide 3+\sqrt{14} by \frac{5}{2} by multiplying 3+\sqrt{14} by the reciprocal of \frac{5}{2}.

$$y=\frac{2\sqrt{14}+6}{5}$$

Now solve the equation y=\frac{3±\sqrt{14}}{\frac{5}{2}} when ± is minus. Subtract \sqrt{14} from 3.

$$y=\frac{3-\sqrt{14}}{\frac{5}{2}}$$

Divide 3-\sqrt{14} by \frac{5}{2} by multiplying 3-\sqrt{14} by the reciprocal of \frac{5}{2}.

$$y=\frac{6-2\sqrt{14}}{5}$$

There are two solutions for y: \frac{6+2\sqrt{14}}{5} and \frac{6-2\sqrt{14}}{5}. Substitute \frac{6+2\sqrt{14}}{5} for y in the equation x=-\frac{1}{2}y+3 to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{1}{2}\times \frac{2\sqrt{14}+6}{5}+3$$

Multiply -\frac{1}{2} times \frac{6+2\sqrt{14}}{5}.

$$x=-\frac{2\sqrt{14}+6}{2\times 5}+3$$

Now substitute \frac{6-2\sqrt{14}}{5} for y in the equation x=-\frac{1}{2}y+3 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{1}{2}\times \frac{6-2\sqrt{14}}{5}+3$$

Multiply -\frac{1}{2} times \frac{6-2\sqrt{14}}{5}.

$$x=-\frac{6-2\sqrt{14}}{2\times 5}+3$$

The system is now solved.

$$x=-\frac{2\sqrt{14}+6}{2\times 5}+3,y=\frac{2\sqrt{14}+6}{5}\text{ or }x=-\frac{6-2\sqrt{14}}{2\times 5}+3,y=\frac{6-2\sqrt{14}}{5}$$