Solve {l}{x^2-4y^2=9}{y=7-3x}

$$\left. \begin{array} { l } { x ^ { 2 } - 4 y ^ { 2 } = 9 } \\ { y = 7 - 3 x } \end{array} \right.$$

Solve for x, y (complex solution)

$$x=\frac{\sqrt{119}i}{35}+\frac{12}{5}\approx 2.4+0.311677489i\text{, }y=-\frac{3\sqrt{119}i}{35}-\frac{1}{5}\approx -0.2-0.935032467i$$

$$x=-\frac{\sqrt{119}i}{35}+\frac{12}{5}\approx 2.4-0.311677489i\text{, }y=\frac{3\sqrt{119}i}{35}-\frac{1}{5}\approx -0.2+0.935032467i$$

Steps Using Substitution and Quadratic Formula

Consider the second equation. Add 3x to both sides.

$$y+3x=7$$

Subtract 3x from both sides of the equation.

$$y=-3x+7$$

Substitute -3x+7 for y in the other equation, x^{2}-4y^{2}=9.

$$x^{2}-4\left(-3x+7\right)^{2}=9$$

Square -3x+7.

$$x^{2}-4\left(9x^{2}-42x+49\right)=9$$

Multiply -4 times 9x^{2}-42x+49.

$$x^{2}-36x^{2}+168x-196=9$$

Add x^{2} to -36x^{2}.

$$-35x^{2}+168x-196=9$$

Subtract 9 from both sides of the equation.

$$-35x^{2}+168x-205=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1-4\left(-3\right)^{2} for a, -4\times 7\left(-3\right)\times 2 for b, and -205 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-168±\sqrt{168^{2}-4\left(-35\right)\left(-205\right)}}{2\left(-35\right)}$$

Square -4\times 7\left(-3\right)\times 2.

$$x=\frac{-168±\sqrt{28224-4\left(-35\right)\left(-205\right)}}{2\left(-35\right)}$$

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

$$x=\frac{-168±\sqrt{28224+140\left(-205\right)}}{2\left(-35\right)}$$

Multiply 140 times -205.

$$x=\frac{-168±\sqrt{28224-28700}}{2\left(-35\right)}$$

Add 28224 to -28700.

$$x=\frac{-168±\sqrt{-476}}{2\left(-35\right)}$$

Take the square root of -476.

$$x=\frac{-168±2\sqrt{119}i}{2\left(-35\right)}$$

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

$$x=\frac{-168±2\sqrt{119}i}{-70}$$

Now solve the equation x=\frac{-168±2\sqrt{119}i}{-70} when ± is plus. Add -168 to 2i\sqrt{119}.

$$x=\frac{-168+2\sqrt{119}i}{-70}$$

Divide -168+2i\sqrt{119} by -70.

$$x=-\frac{\sqrt{119}i}{35}+\frac{12}{5}$$

Now solve the equation x=\frac{-168±2\sqrt{119}i}{-70} when ± is minus. Subtract 2i\sqrt{119} from -168.

$$x=\frac{-2\sqrt{119}i-168}{-70}$$

Divide -168-2i\sqrt{119} by -70.

$$x=\frac{\sqrt{119}i}{35}+\frac{12}{5}$$

There are two solutions for x: \frac{12}{5}-\frac{i\sqrt{119}}{35} and \frac{12}{5}+\frac{i\sqrt{119}}{35}. Substitute \frac{12}{5}-\frac{i\sqrt{119}}{35} for x in the equation y=-3x+7 to find the corresponding solution for y that satisfies both equations.

$$y=-3\left(-\frac{\sqrt{119}i}{35}+\frac{12}{5}\right)+7$$

Now substitute \frac{12}{5}+\frac{i\sqrt{119}}{35} for x in the equation y=-3x+7 and solve to find the corresponding solution for y that satisfies both equations.

$$y=-3\left(\frac{\sqrt{119}i}{35}+\frac{12}{5}\right)+7$$

The system is now solved.

$$y=-3\left(-\frac{\sqrt{119}i}{35}+\frac{12}{5}\right)+7,x=-\frac{\sqrt{119}i}{35}+\frac{12}{5}\text{ or }y=-3\left(\frac{\sqrt{119}i}{35}+\frac{12}{5}\right)+7,x=\frac{\sqrt{119}i}{35}+\frac{12}{5}$$