the sum of two numbers is 11 and the sum of their reciprocal I'd 11/28 find the nos
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Let the number be a and b.
a+b = 11, or
b = 11-a …(1)
i/a + 1/b = 11/3, or
1/b = 11/3 - 1/a …(2)
Multiply (1) by (2), to get
1 = (11-a)(11/3 - 1/a), or
3a = (11-a)(11a-3), or
3a = 121a -11a^2+3a-33, or
11a^2-121a+33 =0 , or
a^2–11a+3=0
a1 = [11+(121–12)^0.5]/2 = [11+109^0.5]/2 =10.72015325 and b1 = 0.279846745
a+b = 11, or
b = 11-a …(1)
i/a + 1/b = 11/3, or
1/b = 11/3 - 1/a …(2)
Multiply (1) by (2), to get
1 = (11-a)(11/3 - 1/a), or
3a = (11-a)(11a-3), or
3a = 121a -11a^2+3a-33, or
11a^2-121a+33 =0 , or
a^2–11a+3=0
a1 = [11+(121–12)^0.5]/2 = [11+109^0.5]/2 =10.72015325 and b1 = 0.279846745
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