Solve {l}{y^2=x^2-64}{3y=x-8}

$$\left. \begin{array} { l } { y ^ { 2 } = x ^ { 2 } - 64 } \\ { 3 y = x - 8 } \end{array} \right.$$

Solve for y, x

$$x=8\text{, }y=0$$

$$x=-10\text{, }y=-6$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Subtract x^{2} from both sides.

$$y^{2}-x^{2}=-64$$

Consider the second equation. Subtract x from both sides.

$$3y-x=-8$$

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.

$$3y-x=-8,-x^{2}+y^{2}=-64$$

Solve 3y-x=-8 for y by isolating y on the left hand side of the equal sign.

$$3y-x=-8$$

Subtract -x from both sides of the equation.

$$3y=x-8$$

Divide both sides by 3.

$$y=\frac{1}{3}x-\frac{8}{3}$$

Substitute \frac{1}{3}x-\frac{8}{3} for y in the other equation, -x^{2}+y^{2}=-64.

$$-x^{2}+\left(\frac{1}{3}x-\frac{8}{3}\right)^{2}=-64$$

Square \frac{1}{3}x-\frac{8}{3}.

$$-x^{2}+\frac{1}{9}x^{2}-\frac{16}{9}x+\frac{64}{9}=-64$$

Add -x^{2} to \frac{1}{9}x^{2}.

$$-\frac{8}{9}x^{2}-\frac{16}{9}x+\frac{64}{9}=-64$$

Add 64 to both sides of the equation.

$$-\frac{8}{9}x^{2}-\frac{16}{9}x+\frac{640}{9}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1+1\times \left(\frac{1}{3}\right)^{2} for a, 1\left(-\frac{8}{3}\right)\times \frac{1}{3}\times 2 for b, and \frac{640}{9} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{16}{9}\right)±\sqrt{\left(-\frac{16}{9}\right)^{2}-4\left(-\frac{8}{9}\right)\times \frac{640}{9}}}{2\left(-\frac{8}{9}\right)}$$

Square 1\left(-\frac{8}{3}\right)\times \frac{1}{3}\times 2.

$$x=\frac{-\left(-\frac{16}{9}\right)±\sqrt{\frac{256}{81}-4\left(-\frac{8}{9}\right)\times \frac{640}{9}}}{2\left(-\frac{8}{9}\right)}$$

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

$$x=\frac{-\left(-\frac{16}{9}\right)±\sqrt{\frac{256}{81}+\frac{32}{9}\times \frac{640}{9}}}{2\left(-\frac{8}{9}\right)}$$

Multiply \frac{32}{9} times \frac{640}{9} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{16}{9}\right)±\sqrt{\frac{256+20480}{81}}}{2\left(-\frac{8}{9}\right)}$$

Add \frac{256}{81} to \frac{20480}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{16}{9}\right)±\sqrt{256}}{2\left(-\frac{8}{9}\right)}$$

Take the square root of 256.

$$x=\frac{-\left(-\frac{16}{9}\right)±16}{2\left(-\frac{8}{9}\right)}$$

The opposite of 1\left(-\frac{8}{3}\right)\times \frac{1}{3}\times 2 is \frac{16}{9}.

$$x=\frac{\frac{16}{9}±16}{2\left(-\frac{8}{9}\right)}$$

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

$$x=\frac{\frac{16}{9}±16}{-\frac{16}{9}}$$

Now solve the equation x=\frac{\frac{16}{9}±16}{-\frac{16}{9}} when ± is plus. Add \frac{16}{9} to 16.

$$x=\frac{\frac{160}{9}}{-\frac{16}{9}}$$

Divide \frac{160}{9} by -\frac{16}{9} by multiplying \frac{160}{9} by the reciprocal of -\frac{16}{9}.

$$x=-10$$

Now solve the equation x=\frac{\frac{16}{9}±16}{-\frac{16}{9}} when ± is minus. Subtract 16 from \frac{16}{9}.

$$x=-\frac{\frac{128}{9}}{-\frac{16}{9}}$$

Divide -\frac{128}{9} by -\frac{16}{9} by multiplying -\frac{128}{9} by the reciprocal of -\frac{16}{9}.

$$x=8$$

There are two solutions for x: -10 and 8. Substitute -10 for x in the equation y=\frac{1}{3}x-\frac{8}{3} to find the corresponding solution for y that satisfies both equations.

$$y=\frac{1}{3}\left(-10\right)-\frac{8}{3}$$

Multiply \frac{1}{3} times -10.

$$y=\frac{-10-8}{3}$$

Add -10\times \frac{1}{3} to -\frac{8}{3}.

$$y=-6$$

Now substitute 8 for x in the equation y=\frac{1}{3}x-\frac{8}{3} and solve to find the corresponding solution for y that satisfies both equations.

$$y=\frac{1}{3}\times 8-\frac{8}{3}$$

Multiply \frac{1}{3} times 8.

$$y=\frac{8-8}{3}$$

Add \frac{1}{3}\times 8 to -\frac{8}{3}.

$$y=0$$

The system is now solved.

$$y=-6,x=-10\text{ or }y=0,x=8$$