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When solving the system of equations y equals x squared, y equals 10 minus three x algebraically, which of the following quadratic equations do you need to solve during the process?
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x2 − 3x − 10 = 0 x2 + 3x − 10 = 0 x2 + 3x + 10 = 0 2y = x2 − 3x + 10
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I think its the last one but not quite sure
That would be incorrect because remember to solve a quadratic it's better if you have all terms on one side
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So what would this system be if you tried to solve it with substitution?
maybe the first no clue honestly
Ok so y=x^2 and y=10-3x So therefore since y=x^2 we can say \[x^2=10-3x\]
Now try and get all the terms on 1 side
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