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anonymous

  • 5 years ago

How to solve system of quadratic linear equations algebraically with y=3x+12 and y=x2-2x-18?

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  1. anonymous
    • 5 years ago
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    substitute y as 3x+12 in the second equation and solve for x.

  2. anonymous
    • 5 years ago
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    Yes sorry, I meant, when you're trying to find out the plotting points, the number pair, what do you do?

  3. anonymous
    • 5 years ago
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    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

  4. anonymous
    • 5 years ago
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    ...Then solve for x, and once you know x, use the simpler equation to find y.

  5. anonymous
    • 5 years ago
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    you find out x and y. then you plotting poitns are (x,y)

  6. anonymous
    • 5 years ago
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    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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