sum of resiprocules of two consicative odd natural number is 12/35 find those number
Answers
Step-by-step explanation:
Let the two consecutive odd natural numbers are x and x+2.
As given sum of reciprocal of two consecutive odd natural number is 12/35 .
Writing the statement in terms of equation:
→1 /x + 1/(x+2) = 12/35
→ \begin{gathered}\frac{x+2+x}{x(x+2)} =\frac{12}{35} \\\frac{2x+2}{x(x+2)} =\frac{12}{35}\end{gathered}
x(x+2)
x+2+x
=
35
12
x(x+2)
2x+2
=
35
12
→35× (2 x +2) = 12× (x² +2 x)
→ 70 x + 70 = 12 x² + 24 x → [ Using Distributive property: a× (b+c) = a ×b + a× c]
Taking variable and constant on one side of equation
→ 12 x² + 24 x - 70 x - 70 =0
→ 12 x² - 46 x - 70=0
→ 2 × (6 x² - 23 x -35) =0
→ 6 x² - 23 x -35=0
Splitting the middle term
→ 6 x² - 30 x + 7 x - 35=0
→ 6 x × ( x -5) + 7 × (x -5) =0
→ (6 x +7)(x-5) =0
→ 6 x +7 =0 ∧ x -5 =0
x ≠ \frac{-7}{6}
6
−7
can't be the solution as x is a natural number. and x = 5.
So, x = 5 and 5+2 =7 are those two numbers whose sum is \frac{12}{35}
35
12
.