Solve {l}{x^2-y^2=16}{3x-5y=0}

$$\left. \begin{array} { l } { x ^ { 2 } - y ^ { 2 } = 16 } \\ { 3 x - 5 y = 0 } \end{array} \right.$$

Solve for x, y

$$x=-5\text{, }y=-3$$

$$x=5\text{, }y=3$$

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.

$$3x-5y=0,-y^{2}+x^{2}=16$$

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

$$3x-5y=0$$

Subtract -5y from both sides of the equation.

$$3x=5y$$

Divide both sides by 3.

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

Substitute \frac{5}{3}y for x in the other equation, -y^{2}+x^{2}=16.

$$-y^{2}+\left(\frac{5}{3}y\right)^{2}=16$$

Square \frac{5}{3}y.

$$-y^{2}+\frac{25}{9}y^{2}=16$$

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

$$\frac{16}{9}y^{2}=16$$

Subtract 16 from both sides of the equation.

$$\frac{16}{9}y^{2}-16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1+1\times \left(\frac{5}{3}\right)^{2} for a, 1\times 0\times \frac{5}{3}\times 2 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{0±\sqrt{0^{2}-4\times \frac{16}{9}\left(-16\right)}}{2\times \frac{16}{9}}$$

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

$$y=\frac{0±\sqrt{-4\times \frac{16}{9}\left(-16\right)}}{2\times \frac{16}{9}}$$

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

$$y=\frac{0±\sqrt{-\frac{64}{9}\left(-16\right)}}{2\times \frac{16}{9}}$$

Multiply -\frac{64}{9} times -16.

$$y=\frac{0±\sqrt{\frac{1024}{9}}}{2\times \frac{16}{9}}$$

Take the square root of \frac{1024}{9}.

$$y=\frac{0±\frac{32}{3}}{2\times \frac{16}{9}}$$

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

$$y=\frac{0±\frac{32}{3}}{\frac{32}{9}}$$

Now solve the equation y=\frac{0±\frac{32}{3}}{\frac{32}{9}} when ± is plus.

$$y=3$$

Now solve the equation y=\frac{0±\frac{32}{3}}{\frac{32}{9}} when ± is minus.

$$y=-3$$

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

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

Multiply \frac{5}{3} times 3.

$$x=5$$

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

$$x=\frac{5}{3}\left(-3\right)$$

Multiply \frac{5}{3} times -3.

$$x=-5$$

The system is now solved.

$$x=5,y=3\text{ or }x=-5,y=-3$$