Chapter 9 - 9.1 - Linear and Nonlinear Systems of Equations - 9.1 Exercise - Page 636: 30

$(0,0),$ $(12,6)$

From the first equation, $x=2y$ (Isolation). Consider the second equation, $3x-y^{2}=0$ This becomes: $3(2y)-y^{2}=0$ (Substitution). Thus, $6y-y^{2}=0$ $y(6-y)=0$ This gives $y=0,$ and $y=6$ Accordingly, $x=0,$ and $x=12$