Question 1058629
 2. Two numbers 
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x = first number
y = second number
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are such that if the square of the first number 
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The square of the first number is x<sup>2</sup>
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is subtracted from twice their product, 
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Twice their product is 2xy

So we subtract x<sup>2</sup> from 2xy

That's 2xy - x<sup>2</sup>
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the difference is -1. 
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So we set that difference (subtraction) equal to -1 
and that's our first equation:

2xy - x<sup>2</sup> = -1
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But twice the product 
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That's xy
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added to the sum of thrice the square of the first number 
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that's 3x<sup>2</sup>
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and five times that number 
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that's 5x

So the sum of those is 3x<sup>2</sup> + 5x
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gives 10
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So we add xy to that, which is 3x<sup>2</sup> + 5x + xy 

and we set that = 10, and the second equation is

3x<sup>2</sup> + 5x + xy = 10 

So the system of equations is

{{{system(2xy - x^2 = -1, 3x^2 + 5x + xy = 10)}}} 

That has solutions:

{{{x = (-5 - 2sqrt(43))/7}}}, {{{y = (-10 - 2sqrt(43))/21}}}

and

{{{x = (-5 + 2sqrt(43))/7}}}, {{{y = (-10 + 2sqrt(43))/21}}}


Edwin</pre>