How to solve system of quadratic linear equations algebraically with y=3x+12 and y=x2-2x-18?
substitute y as 3x+12 in the second equation and solve for x.
Yes sorry, I meant, when you're trying to find out the plotting points, the number pair, what do you do?
They are both written as y = ... So knowing that y = y, the other sides are equal to each other too, so set the other sides equal to each other: 3x + 12 = x^2-2x-18
...Then solve for x, and once you know x, use the simpler equation to find y.
you find out x and y. then you plotting poitns are (x,y)
I don't know what you mean, because there is only only point (x,y) or at most two, which satisfies both these equations. You have to find that point by substituting the first equation in the second equation.
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