Guest Posted October 28, 2011 Report Share Posted October 28, 2011 How many pairs of real numbers (x,y) satisfy the equation...... ? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted October 28, 2011 Report Share Posted October 28, 2011 x = (-3) y = 3 Quote Link to comment Share on other sites More sharing options...
0 hhh3 Posted October 28, 2011 Report Share Posted October 28, 2011 well... now i would say 1. all thanks to SMV Quote Link to comment Share on other sites More sharing options...
0 Guest Posted October 28, 2011 Report Share Posted October 28, 2011 By doing the multiplication and collecting terms, we can see that: x^2 +(Y+3)x+(y^2-3y+9)=0 or if you prefer y^2 + (x-3)y+(x^2+3X+9)=0 Using the quadratic formula, we observe that the term (b^2-4ac) must be greater than or equal to 0 for there to be a real solution. For the first equation, this becomes: y^2-6y+9 >=0 which is true for Y=3 For the latter equation, this becomes: x^2+6x+9>=0 which is true for x=-3 Both of these equations must be true and (-3.3) is the only solution that satisfies both. Hence the answer is one pair. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted October 28, 2011 Report Share Posted October 28, 2011 I'm sure he thanks you for doing his math homework for him Quote Link to comment Share on other sites More sharing options...
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How many pairs of real numbers (x,y) satisfy the equation...... ?
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