Look at the system. You see that the two terms with x^2 and x of the second equation are five times
the corresponding terms of the first equation.
It gives me an idea to multiply first equation by 5 and then subtract from the second equation
to ELIMINATE all the terms with x.
So my first step is to multiply first equation by 5, while keeping second equation as is
5x^2 - 10x - 20y^2 + 40y - 10 = 0 (1')
5x^2 - 10x + 12y^2 + 24y - 58 = 0 (2)
My second step is to subtract equation (1') from equation (2). In this way, I get
32y^2 - 16y - 48 = 0.
Next, use the quadratic formula
= = = .
So, I have two solutions for y: = -1 and = = .
The last step is to substitute these found values of y into equation (1) and solve it for x.
It is just fully mechanical job and I leave it to you.
You should find two values of x for each of the two values of y.
It will give you four final solutions for pairs (x,y) at the end.
Having so clear instructions, boldly go forward and complete the solution on your own.
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By the way, first equation represents a hyperbola, while the second equation represents an ellipse.
The solution represents 4 (four) intersection points.
It is good if you understand what these equations represent.
It is not necessary for formal solution at this level, but nevertheless . . .
