1. **State the problem:** We have the equation $$2x^2 - 3xy - y^2 + x - 5y - 3 = 0$$ and want to translate the origin to a point $ (h,k) $ so that the new equation has no first degree terms in $x$ or $y$. 2. **Formula and approach:** When translating the origin to $ (h,k) $, substitute $$x = X + h$$ and $$y = Y + k$$ into the equation. The new equation in terms of $X$ and $Y$ will have no linear terms if the coefficients of $X$ and $Y$ vanish. 3. **Substitute and expand:** $$2(X+h)^2 - 3(X+h)(Y+k) - (Y+k)^2 + (X+h) - 5(Y+k) - 3 = 0$$ Expanding: $$2(X^2 + 2hX + h^2) - 3(XY + Xk + hY + hk) - (Y^2 + 2kY + k^2) + X + h - 5Y - 5k - 3 = 0$$ 4. **Group terms:** Quadratic terms: $$2X^2 - 3XY - Y^2$$ Linear terms: $$2 imes 2hX - 3(Xk + hY) + X - 5Y = (4h - 3k + 1)X + (-3h - 2k - 5)Y$$ Constant terms: $$2h^2 - 3hk - k^2 + h - 5k - 3$$ 5. **Set linear coefficients to zero:** $$4h - 3k + 1 = 0$$ $$-3h - 2k - 5 = 0$$ 6. **Solve the system:** From the first: $$4h = 3k - 1$$ $$h = \frac{3k - 1}{4}$$ Substitute into second: $$-3\left(\frac{3k - 1}{4}\right) - 2k - 5 = 0$$ Multiply both sides by 4: $$-3(3k - 1) - 8k - 20 = 0$$ $$-9k + 3 - 8k - 20 = 0$$ $$-17k - 17 = 0$$ $$-17k = 17$$ $$k = -1$$ Substitute back: $$h = \frac{3(-1) - 1}{4} = \frac{-3 - 1}{4} = \frac{-4}{4} = -1$$ 7. **Answer:** The new origin is at $$\boxed{(-1, -1)}$$ where the equation has no first degree terms.