How to solve the equation system x square + 4x + 4Y square = 9; X square + xy = 0

x^+4x+4y^=9 ...(1)
x^+xy=0 ...(2)
For (1), x = 0, y = 1.5 or - 1.5
When x is not 0, we substitute x = - y from (2) into 1 to get 5Y ^ - 4y-9 = 0, and the solution is y = - 1 or y = 1.8
So there are four sets of solutions
x=0,y=1.5;
x=0,y=-1.5;
x=1,y=-1;
x=-1.8,y=1.8;
Hope to help you

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