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

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

Solve for x, y (complex solution)

$$x=-1+2i\text{, }y=2-3i$$

$$x=-1-2i\text{, }y=2+3i$$

Steps Using Substitution and Quadratic Formula

Solve 3x+2y=1 for x by isolating x on the left hand side of the equal sign.

$$3x+2y=1$$

Subtract 2y from both sides of the equation.

$$3x=-2y+1$$

Divide both sides by 3.

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

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

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

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

$$-y^{2}+3\left(\frac{4}{9}y^{2}-\frac{4}{9}y+\frac{1}{9}\right)=-4$$

Multiply 3 times \frac{4}{9}y^{2}-\frac{4}{9}y+\frac{1}{9}.

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

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

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

Add 4 to both sides of the equation.

$$\frac{1}{3}y^{2}-\frac{4}{3}y+\frac{13}{3}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1+3\left(-\frac{2}{3}\right)^{2} for a, 3\times \frac{1}{3}\left(-\frac{2}{3}\right)\times 2 for b, and \frac{13}{3} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-\frac{4}{3}\right)±\sqrt{\left(-\frac{4}{3}\right)^{2}-4\times \frac{1}{3}\times \frac{13}{3}}}{2\times \frac{1}{3}}$$

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

$$y=\frac{-\left(-\frac{4}{3}\right)±\sqrt{\frac{16}{9}-4\times \frac{1}{3}\times \frac{13}{3}}}{2\times \frac{1}{3}}$$

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

$$y=\frac{-\left(-\frac{4}{3}\right)±\sqrt{\frac{16}{9}-\frac{4}{3}\times \frac{13}{3}}}{2\times \frac{1}{3}}$$

Multiply -\frac{4}{3} times \frac{13}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{4}{3}\right)±\sqrt{\frac{16-52}{9}}}{2\times \frac{1}{3}}$$

Add \frac{16}{9} to -\frac{52}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Take the square root of -4.

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

The opposite of 3\times \frac{1}{3}\left(-\frac{2}{3}\right)\times 2 is \frac{4}{3}.

$$y=\frac{\frac{4}{3}±2i}{2\times \frac{1}{3}}$$

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

$$y=\frac{\frac{4}{3}±2i}{\frac{2}{3}}$$

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

$$y=\frac{\frac{4}{3}+2i}{\frac{2}{3}}$$

Divide \frac{4}{3}+2i by \frac{2}{3} by multiplying \frac{4}{3}+2i by the reciprocal of \frac{2}{3}.

$$y=2+3i$$

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

$$y=\frac{\frac{4}{3}-2i}{\frac{2}{3}}$$

Divide \frac{4}{3}-2i by \frac{2}{3} by multiplying \frac{4}{3}-2i by the reciprocal of \frac{2}{3}.

$$y=2-3i$$

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

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

Multiply -\frac{2}{3} times 2+3i.

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

Add -\frac{2}{3}\left(2+3i\right) to \frac{1}{3}.

$$x=-1-2i$$

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

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

Multiply -\frac{2}{3} times 2-3i.

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

Add -\frac{2}{3}\left(2-3i\right) to \frac{1}{3}.

$$x=-1+2i$$

The system is now solved.

$$x=-1-2i,y=2+3i\text{ or }x=-1+2i,y=2-3i$$