Divide 15 into two parts such that the sum of its reciprocal is 3/10
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Answer:
Step-by-step explanation:
Let first part be 'x' and second part be '15-x'.
There reciprocals are (1/x) and 1/(15-x) respectively.
According to question-
(1/x) + 1/(15-x) = 3/10
(15-x+x)/[x(15-x)]=3/10
15/(15x-x^2)=3/10
45x-3x^2=150
taking 150 to LHS and Dividing each side by -3
x^2-15x+50=0
This can be written as-
x^2-5x-10x+50=0
x(x-5)-10(x-5)=0
(x-5)(x-10)=0
x=5 and 10
Therefore required numbers are 5 and 10
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