Solve {l}{x+2y=10}{3x^2-y^2=11}

$$\left\{ \begin{array} { l } { x + 2 y = 10 } \\ { 3 x ^ { 2 } - y ^ { 2 } = 11 } \end{array} \right.$$

Solve for x, y

$$x=\frac{2\sqrt{421}-10}{11}\approx 2.821506278\text{, }y=\frac{60-\sqrt{421}}{11}\approx 3.589246861$$

$$x=\frac{-2\sqrt{421}-10}{11}\approx -4.639688096\text{, }y=\frac{\sqrt{421}+60}{11}\approx 7.319844048$$

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.

$$x+2y=10,-y^{2}+3x^{2}=11$$

Solve x+2y=10 for x by isolating x on the left hand side of the equal sign.

$$x+2y=10$$

Subtract 2y from both sides of the equation.

$$x=-2y+10$$

Substitute -2y+10 for x in the other equation, -y^{2}+3x^{2}=11.

$$-y^{2}+3\left(-2y+10\right)^{2}=11$$

Square -2y+10.

$$-y^{2}+3\left(4y^{2}-40y+100\right)=11$$

Multiply 3 times 4y^{2}-40y+100.

$$-y^{2}+12y^{2}-120y+300=11$$

Add -y^{2} to 12y^{2}.

$$11y^{2}-120y+300=11$$

Subtract 11 from both sides of the equation.

$$11y^{2}-120y+289=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1+3\left(-2\right)^{2} for a, 3\times 10\left(-2\right)\times 2 for b, and 289 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-120\right)±\sqrt{\left(-120\right)^{2}-4\times 11\times 289}}{2\times 11}$$

Square 3\times 10\left(-2\right)\times 2.

$$y=\frac{-\left(-120\right)±\sqrt{14400-4\times 11\times 289}}{2\times 11}$$

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

$$y=\frac{-\left(-120\right)±\sqrt{14400-44\times 289}}{2\times 11}$$

Multiply -44 times 289.

$$y=\frac{-\left(-120\right)±\sqrt{14400-12716}}{2\times 11}$$

Add 14400 to -12716.

$$y=\frac{-\left(-120\right)±\sqrt{1684}}{2\times 11}$$

Take the square root of 1684.

$$y=\frac{-\left(-120\right)±2\sqrt{421}}{2\times 11}$$

The opposite of 3\times 10\left(-2\right)\times 2 is 120.

$$y=\frac{120±2\sqrt{421}}{2\times 11}$$

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

$$y=\frac{120±2\sqrt{421}}{22}$$

Now solve the equation y=\frac{120±2\sqrt{421}}{22} when ± is plus. Add 120 to 2\sqrt{421}.

$$y=\frac{2\sqrt{421}+120}{22}$$

Divide 120+2\sqrt{421} by 22.

$$y=\frac{\sqrt{421}+60}{11}$$

Now solve the equation y=\frac{120±2\sqrt{421}}{22} when ± is minus. Subtract 2\sqrt{421} from 120.

$$y=\frac{120-2\sqrt{421}}{22}$$

Divide 120-2\sqrt{421} by 22.

$$y=\frac{60-\sqrt{421}}{11}$$

There are two solutions for y: \frac{60+\sqrt{421}}{11} and \frac{60-\sqrt{421}}{11}. Substitute \frac{60+\sqrt{421}}{11} for y in the equation x=-2y+10 to find the corresponding solution for x that satisfies both equations.

$$x=-2\times \frac{\sqrt{421}+60}{11}+10$$

Now substitute \frac{60-\sqrt{421}}{11} for y in the equation x=-2y+10 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-2\times \frac{60-\sqrt{421}}{11}+10$$

The system is now solved.

$$x=-2\times \frac{\sqrt{421}+60}{11}+10,y=\frac{\sqrt{421}+60}{11}\text{ or }x=-2\times \frac{60-\sqrt{421}}{11}+10,y=\frac{60-\sqrt{421}}{11}$$