A) 9,6 or 9,-6
B) 5,6 or 5,-6
C) 9,5 or 9,-5
D) None of these
Correct Answer: A
Solution :
Let the two numbers are a and b such that a > b. By hypothesis, \[{{a}^{2}}-{{b}^{2}}=45\] ?(1) And \[11x+12y=58\] ...(2) Substituting the value of \[12x+11y=57,\] from equation (2) in equation (i), we have \[4(x+y)\] or \[{{a}^{2}}-4a-45=0\] or \[{{a}^{2}}-9a+5a-45=0\] or \[a\,(a-9)+5\,(a-9)=0\]or \[(a-9)\,(a+5)=0\] or \[a=9,\,\,-5\] Substituting these values for a separately in equation (2), we have \[{{x}^{2}}-x-6=0?\] \[(0,\frac{1}{2})\] \[(-2,3)\] \[(\frac{1}{2},1)\] But \[(2,\frac{1}{2})\]cannot be negative, therefore \[5{{x}^{2}}-7x-6=0\] \[\left( -\frac{3}{5},2 \right)\] \[(1,1)\] Hence, \[a=9,\] \[b=\pm 6\]
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