Solve {l}{x^2-y^2=25}{5x-3y=0}

$$\left. \begin{array} { l } { x ^ { 2 } - y ^ { 2 } = 25 } \\ { 5 x - 3 y = 0 } \end{array} \right.$$

Solve for x, y (complex solution)

$$x=-\frac{15}{4}i=-3.75i\text{, }y=-\frac{25}{4}i=-6.25i$$

$$x=\frac{15}{4}i=3.75i\text{, }y=\frac{25}{4}i=6.25i$$

Steps Using Substitution and Quadratic Formula

Solve 5x-3y=0 for x by isolating x on the left hand side of the equal sign.

$$5x-3y=0$$

Subtract -3y from both sides of the equation.

$$5x=3y$$

Divide both sides by 5.

$$x=\frac{3}{5}y$$

Substitute \frac{3}{5}y for x in the other equation, -y^{2}+x^{2}=25.

$$-y^{2}+\left(\frac{3}{5}y\right)^{2}=25$$

Square \frac{3}{5}y.

$$-y^{2}+\frac{9}{25}y^{2}=25$$

Add -y^{2} to \frac{9}{25}y^{2}.

$$-\frac{16}{25}y^{2}=25$$

Subtract 25 from both sides of the equation.

$$-\frac{16}{25}y^{2}-25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1+1\times \left(\frac{3}{5}\right)^{2} for a, 1\times 0\times \frac{3}{5}\times 2 for b, and -25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{0±\sqrt{0^{2}-4\left(-\frac{16}{25}\right)\left(-25\right)}}{2\left(-\frac{16}{25}\right)}$$

Square 1\times 0\times \frac{3}{5}\times 2.

$$y=\frac{0±\sqrt{-4\left(-\frac{16}{25}\right)\left(-25\right)}}{2\left(-\frac{16}{25}\right)}$$

Multiply -4 times -1+1\times \left(\frac{3}{5}\right)^{2}.

$$y=\frac{0±\sqrt{\frac{64}{25}\left(-25\right)}}{2\left(-\frac{16}{25}\right)}$$

Multiply \frac{64}{25} times -25.

$$y=\frac{0±\sqrt{-64}}{2\left(-\frac{16}{25}\right)}$$

Take the square root of -64.

$$y=\frac{0±8i}{2\left(-\frac{16}{25}\right)}$$

Multiply 2 times -1+1\times \left(\frac{3}{5}\right)^{2}.

$$y=\frac{0±8i}{-\frac{32}{25}}$$

Now solve the equation y=\frac{0±8i}{-\frac{32}{25}} when ± is plus.

$$y=-\frac{25}{4}i$$

Now solve the equation y=\frac{0±8i}{-\frac{32}{25}} when ± is minus.

$$y=\frac{25}{4}i$$

There are two solutions for y: -\frac{25}{4}i and \frac{25}{4}i. Substitute -\frac{25}{4}i for y in the equation x=\frac{3}{5}y to find the corresponding solution for x that satisfies both equations.

$$x=\frac{3}{5}\times \left(-\frac{25}{4}i\right)$$

Multiply \frac{3}{5} times -\frac{25}{4}i.

$$x=-\frac{15}{4}i$$

Now substitute \frac{25}{4}i for y in the equation x=\frac{3}{5}y and solve to find the corresponding solution for x that satisfies both equations.

$$x=\frac{3}{5}\times \left(\frac{25}{4}i\right)$$

Multiply \frac{3}{5} times \frac{25}{4}i.

$$x=\frac{15}{4}i$$

The system is now solved.

$$x=-\frac{15}{4}i,y=-\frac{25}{4}i\text{ or }x=\frac{15}{4}i,y=\frac{25}{4}i$$