To solve the system of equations, we can use substitution or elimination. Here, we used substitution. First, we solved the second equation for x in terms of y, yielding x = (6 - 2y)/3. We then substituted this expression for x into the first equation, resulting in a quadratic equation in terms of y. After simplifying, we factored the equation to find the possible values for y: y = 0 and y = 3. We then substituted these values back into the second equation to find the corresponding x values, leading to the solutions (2, 0) and (0, 3). This method illustrates how to manipulate equations to find the intersection points of two lines represented by the equations.