The sum of reciprocals of two consecutive numbers is 23/132. Find the numbers.
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Let the Consecutive be x and x+1
Their reciprocal be 1/x+1/x+1
Their Sum = 23/132
1/x+1/x+1=23/132
LCM of x and x+1
1(x+1)+1(x)/x(x+1)=23/132
x+1+x/x²+x=23/132
2x+1/x²+x=23/132
Cross multiply
2x+1(132)=23(x²+x)
264x+132=23x²+23x
23x+23x-264x-132
23x-241x-132
(Factorise them and u will get ur answer)
Their reciprocal be 1/x+1/x+1
Their Sum = 23/132
1/x+1/x+1=23/132
LCM of x and x+1
1(x+1)+1(x)/x(x+1)=23/132
x+1+x/x²+x=23/132
2x+1/x²+x=23/132
Cross multiply
2x+1(132)=23(x²+x)
264x+132=23x²+23x
23x+23x-264x-132
23x-241x-132
(Factorise them and u will get ur answer)
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