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An example of a non-linear equation is a quadratic equation of the form ax^2 + bx + c = 0

To get two inputs you will have to equations like this

y = x^2

y = 8 - x^2

You need to find the points where the equations overlap, so you are trying to find any spots where y = x^2 equals y = 8 - x^2:

y = x^2 = y = 8 - x^2

The algebra comes in when I manipulate useful bits of this last equation. I can pick out whichever parts I like - they're all equal, after all

y = x2

y = 8 - x2

y = y

x^2 = 8 - x^2

Each of these sub-equations is true, but only the last one is usefully new and different. I can solve this for the x-values that make the equation true:

x^2 = 8 - x^2

2x^2 = 8

x^2 = 4

x = -2, +2

Then the solutions to the original system will occur when x = -2 and when x = +2.

What are the corresponding y-values? To find them, I plug the x-values back in to either original equation. (It doesn't matter which one I pick because I only care about the points where the equations spit out the same values. So I can pick whichever equation I like better.) I'll plug the x-values into the first equation, because it's the simpler of the two:

x = -2:

y = x^2

y = (-2)^2 = 4

x = +2:

y = x^2

y = (+2)^2 = 4

Then the solutions (as we already knew) are (x, y) = (-2, 4) and (2, 4).

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