Chapter 9 Quadratic Relations and Conic Sections - Chapter Review - Page 672: 37

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$16x^2-4y^2=64\\4x^2+9y^2-40x=-64$ Multiply the first equation by 9 and the second equation by 4, and then add them together: $144x^2-36y^2+16x^2+36y^2-160x=576-256\\160x^2-160x-320=0\\x^2-x-2=0$ Thus $x=-1$ or $x=2$. Thus if $x=-1$, then there is no real solution, and if $x=2$, then $y=0$. Thus the solution is: $(2,0)$