Solve {l}{x-y=1/4}{3x^2-6=y^2}

$$\left\{ \begin{array} { l } { x - y = \frac { 1 } { 4 } } \\ { 3 x ^ { 2 } - 6 = y ^ { 2 } } \end{array} \right.$$

Solve for x, y

$$x=\frac{-\sqrt{195}-1}{8}\approx -1.870530005\text{, }y=\frac{-\sqrt{195}-3}{8}\approx -2.120530005$$

$$x=\frac{\sqrt{195}-1}{8}\approx 1.620530005\text{, }y=\frac{\sqrt{195}-3}{8}\approx 1.370530005$$

Steps Using Substitution and Quadratic Formula

Consider the second equation. Subtract y^{2} from both sides.

$$3x^{2}-6-y^{2}=0$$

Add 6 to both sides. Anything plus zero gives itself.

$$3x^{2}-y^{2}=6$$

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.

$$x-y=\frac{1}{4},-y^{2}+3x^{2}=6$$

Solve x-y=\frac{1}{4} for x by isolating x on the left hand side of the equal sign.

$$x-y=\frac{1}{4}$$

Subtract -y from both sides of the equation.

$$x=y+\frac{1}{4}$$

Substitute y+\frac{1}{4} for x in the other equation, -y^{2}+3x^{2}=6.

$$-y^{2}+3\left(y+\frac{1}{4}\right)^{2}=6$$

Square y+\frac{1}{4}.

$$-y^{2}+3\left(y^{2}+\frac{1}{2}y+\frac{1}{16}\right)=6$$

Multiply 3 times y^{2}+\frac{1}{2}y+\frac{1}{16}.

$$-y^{2}+3y^{2}+\frac{3}{2}y+\frac{3}{16}=6$$

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

$$2y^{2}+\frac{3}{2}y+\frac{3}{16}=6$$

Subtract 6 from both sides of the equation.

$$2y^{2}+\frac{3}{2}y-\frac{93}{16}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1+3\times 1^{2} for a, 3\times \frac{1}{4}\times 1\times 2 for b, and -\frac{93}{16} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\frac{3}{2}±\sqrt{\left(\frac{3}{2}\right)^{2}-4\times 2\left(-\frac{93}{16}\right)}}{2\times 2}$$

Square 3\times \frac{1}{4}\times 1\times 2.

$$y=\frac{-\frac{3}{2}±\sqrt{\frac{9}{4}-4\times 2\left(-\frac{93}{16}\right)}}{2\times 2}$$

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

$$y=\frac{-\frac{3}{2}±\sqrt{\frac{9}{4}-8\left(-\frac{93}{16}\right)}}{2\times 2}$$

Multiply -8 times -\frac{93}{16}.

$$y=\frac{-\frac{3}{2}±\sqrt{\frac{9}{4}+\frac{93}{2}}}{2\times 2}$$

Add \frac{9}{4} to \frac{93}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\frac{3}{2}±\sqrt{\frac{195}{4}}}{2\times 2}$$

Take the square root of \frac{195}{4}.

$$y=\frac{-\frac{3}{2}±\frac{\sqrt{195}}{2}}{2\times 2}$$

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

$$y=\frac{-\frac{3}{2}±\frac{\sqrt{195}}{2}}{4}$$

Now solve the equation y=\frac{-\frac{3}{2}±\frac{\sqrt{195}}{2}}{4} when ± is plus. Add -\frac{3}{2} to \frac{\sqrt{195}}{2}.

$$y=\frac{\sqrt{195}-3}{2\times 4}$$

Divide \frac{-3+\sqrt{195}}{2} by 4.

$$y=\frac{\sqrt{195}-3}{8}$$

Now solve the equation y=\frac{-\frac{3}{2}±\frac{\sqrt{195}}{2}}{4} when ± is minus. Subtract \frac{\sqrt{195}}{2} from -\frac{3}{2}.

$$y=\frac{-\sqrt{195}-3}{2\times 4}$$

Divide \frac{-3-\sqrt{195}}{2} by 4.

$$y=\frac{-\sqrt{195}-3}{8}$$

There are two solutions for y: \frac{-3+\sqrt{195}}{8} and \frac{-3-\sqrt{195}}{8}. Substitute \frac{-3+\sqrt{195}}{8} for y in the equation x=y+\frac{1}{4} to find the corresponding solution for x that satisfies both equations.

$$x=\frac{\sqrt{195}-3}{8}+\frac{1}{4}$$

Now substitute \frac{-3-\sqrt{195}}{8} for y in the equation x=y+\frac{1}{4} and solve to find the corresponding solution for x that satisfies both equations.

$$x=\frac{-\sqrt{195}-3}{8}+\frac{1}{4}$$

The system is now solved.

$$x=\frac{\sqrt{195}-3}{8}+\frac{1}{4},y=\frac{\sqrt{195}-3}{8}\text{ or }x=\frac{-\sqrt{195}-3}{8}+\frac{1}{4},y=\frac{-\sqrt{195}-3}{8}$$