Solve {l}{x^2/6+y^2/4=1}{x=y+1}

$$\left\{ \begin{array} { l } { \frac { x ^ { 2 } } { 6 } + \frac { y ^ { 2 } } { 4 } = 1 } \\ { x = y + 1 } \end{array} \right.$$

Solve for x, y

$$x=\frac{3-3\sqrt{6}}{5}\approx -0.869693846\text{, }y=\frac{-3\sqrt{6}-2}{5}\approx -1.869693846$$

$$x=\frac{3\sqrt{6}+3}{5}\approx 2.069693846\text{, }y=\frac{3\sqrt{6}-2}{5}\approx 1.069693846$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Multiply both sides of the equation by 12, the least common multiple of 6,4.

$$2x^{2}+3y^{2}=12$$

Consider the second equation. Subtract y from both sides.

$$x-y=1$$

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.

$$x-y=1,3y^{2}+2x^{2}=12$$

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

$$x-y=1$$

Subtract -y from both sides of the equation.

$$x=y+1$$

Substitute y+1 for x in the other equation, 3y^{2}+2x^{2}=12.

$$3y^{2}+2\left(y+1\right)^{2}=12$$

Square y+1.

$$3y^{2}+2\left(y^{2}+2y+1\right)=12$$

Multiply 2 times y^{2}+2y+1.

$$3y^{2}+2y^{2}+4y+2=12$$

Add 3y^{2} to 2y^{2}.

$$5y^{2}+4y+2=12$$

Subtract 12 from both sides of the equation.

$$5y^{2}+4y-10=0$$

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

$$y=\frac{-4±\sqrt{4^{2}-4\times 5\left(-10\right)}}{2\times 5}$$

Square 2\times 1\times 1\times 2.

$$y=\frac{-4±\sqrt{16-4\times 5\left(-10\right)}}{2\times 5}$$

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

$$y=\frac{-4±\sqrt{16-20\left(-10\right)}}{2\times 5}$$

Multiply -20 times -10.

$$y=\frac{-4±\sqrt{16+200}}{2\times 5}$$

Add 16 to 200.

$$y=\frac{-4±\sqrt{216}}{2\times 5}$$

Take the square root of 216.

$$y=\frac{-4±6\sqrt{6}}{2\times 5}$$

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

$$y=\frac{-4±6\sqrt{6}}{10}$$

Now solve the equation y=\frac{-4±6\sqrt{6}}{10} when ± is plus. Add -4 to 6\sqrt{6}.

$$y=\frac{6\sqrt{6}-4}{10}$$

Divide -4+6\sqrt{6} by 10.

$$y=\frac{3\sqrt{6}-2}{5}$$

Now solve the equation y=\frac{-4±6\sqrt{6}}{10} when ± is minus. Subtract 6\sqrt{6} from -4.

$$y=\frac{-6\sqrt{6}-4}{10}$$

Divide -4-6\sqrt{6} by 10.

$$y=\frac{-3\sqrt{6}-2}{5}$$

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

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

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

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

The system is now solved.

$$x=\frac{3\sqrt{6}-2}{5}+1,y=\frac{3\sqrt{6}-2}{5}\text{ or }x=\frac{-3\sqrt{6}-2}{5}+1,y=\frac{-3\sqrt{6}-2}{5}$$