Solve {l}{3x^2-y^2=0}{2y+x=2}

$$\left\{ \begin{array} { l } { 3 x ^ { 2 } - y ^ { 2 } = 0 } \\ { 2 y + x = 2 } \end{array} \right.$$

Solve for x, y

$$x=\frac{4\sqrt{3}-2}{11}\approx 0.448018475\text{, }y=\frac{12-2\sqrt{3}}{11}\approx 0.775990762$$

$$x=\frac{-4\sqrt{3}-2}{11}\approx -0.811654839\text{, }y=\frac{2\sqrt{3}+12}{11}\approx 1.40582742$$

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.

$$2y+x=2,3x^{2}-y^{2}=0$$

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

$$2y+x=2$$

Subtract x from both sides of the equation.

$$2y=-x+2$$

Divide both sides by 2.

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

Substitute -\frac{1}{2}x+1 for y in the other equation, 3x^{2}-y^{2}=0.

$$3x^{2}-\left(-\frac{1}{2}x+1\right)^{2}=0$$

Square -\frac{1}{2}x+1.

$$3x^{2}-\left(\frac{1}{4}x^{2}-x+1\right)=0$$

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

$$3x^{2}-\frac{1}{4}x^{2}+x-1=0$$

Add 3x^{2} to -\frac{1}{4}x^{2}.

$$\frac{11}{4}x^{2}+x-1=0$$

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

$$x=\frac{-1±\sqrt{1^{2}-4\times \frac{11}{4}\left(-1\right)}}{2\times \frac{11}{4}}$$

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

$$x=\frac{-1±\sqrt{1-4\times \frac{11}{4}\left(-1\right)}}{2\times \frac{11}{4}}$$

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

$$x=\frac{-1±\sqrt{1-11\left(-1\right)}}{2\times \frac{11}{4}}$$

Multiply -11 times -1.

$$x=\frac{-1±\sqrt{1+11}}{2\times \frac{11}{4}}$$

Add 1 to 11.

$$x=\frac{-1±\sqrt{12}}{2\times \frac{11}{4}}$$

Take the square root of 12.

$$x=\frac{-1±2\sqrt{3}}{2\times \frac{11}{4}}$$

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

$$x=\frac{-1±2\sqrt{3}}{\frac{11}{2}}$$

Now solve the equation x=\frac{-1±2\sqrt{3}}{\frac{11}{2}} when ± is plus. Add -1 to 2\sqrt{3}.

$$x=\frac{2\sqrt{3}-1}{\frac{11}{2}}$$

Divide -1+2\sqrt{3} by \frac{11}{2} by multiplying -1+2\sqrt{3} by the reciprocal of \frac{11}{2}.

$$x=\frac{4\sqrt{3}-2}{11}$$

Now solve the equation x=\frac{-1±2\sqrt{3}}{\frac{11}{2}} when ± is minus. Subtract 2\sqrt{3} from -1.

$$x=\frac{-2\sqrt{3}-1}{\frac{11}{2}}$$

Divide -1-2\sqrt{3} by \frac{11}{2} by multiplying -1-2\sqrt{3} by the reciprocal of \frac{11}{2}.

$$x=\frac{-4\sqrt{3}-2}{11}$$

There are two solutions for x: \frac{-2+4\sqrt{3}}{11} and \frac{-2-4\sqrt{3}}{11}. Substitute \frac{-2+4\sqrt{3}}{11} for x in the equation y=-\frac{1}{2}x+1 to find the corresponding solution for y that satisfies both equations.

$$y=-\frac{1}{2}\times \frac{4\sqrt{3}-2}{11}+1$$

Multiply -\frac{1}{2} times \frac{-2+4\sqrt{3}}{11}.

$$y=-\frac{4\sqrt{3}-2}{2\times 11}+1$$

Now substitute \frac{-2-4\sqrt{3}}{11} for x in the equation y=-\frac{1}{2}x+1 and solve to find the corresponding solution for y that satisfies both equations.

$$y=-\frac{1}{2}\times \frac{-4\sqrt{3}-2}{11}+1$$

Multiply -\frac{1}{2} times \frac{-2-4\sqrt{3}}{11}.

$$y=-\frac{-4\sqrt{3}-2}{2\times 11}+1$$

The system is now solved.

$$y=-\frac{4\sqrt{3}-2}{2\times 11}+1,x=\frac{4\sqrt{3}-2}{11}\text{ or }y=-\frac{-4\sqrt{3}-2}{2\times 11}+1,x=\frac{-4\sqrt{3}-2}{11}$$