Question

the sum of the reciprocals of two consecutive odd number is 12/35. find those numbers

Answer 1

Here your answer goes

Step :- 1

Let two Consecutive odd integer be x and x +2

Step :- 2

According to Question :-

$\frac{1}{x} + \frac{1}{x +2} = \frac{12}{35}$

⇒ $35(x+2) + 35x = 12x(x+2)$

⇒ $35x + 70 + 35x = 12x^2 +24$

⇒ $70x +70 = 12x^2 +24$

⇒ $12x^2 +24x - 70x - 70 = 0$

⇒ $12x^2 + 46x - 70 = 0$

⇒ x =  5 and $\frac{-7}{6}$

We neglect $\frac{-7}{6}$

So , x = 5

x +2

5 + 2

= 7

Therefore , two numbers is 5 and 7

Verification :-

$\frac{1}{5} + \frac{1}{7}$

⇒ $\frac{7+5}{35}$

⇒ $\frac{12}{35}$

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Answer 2

нєуα

нєяє ιѕ тнє αиѕωєя

ℓєт тнє иυмвєяѕ вє χ αи∂ χ + 2.

α/q

$\frac{1}{x} + \frac{1}{x + 2} = \frac{12}{35} \\ \\ = > \frac{x + 2 + x}{(x)(x + 2)} = \frac{12}{35} \\ \\ = > \frac{2 + 2x}{ {x}^{2} + 2x } = \frac{12}{35} \\ \\ cross \: multiplication \\ \\ = > 35(2 + 2x) = 12( {x}^{2} + 2x) \\ \\ = > 70 + 70x = 12 {x}^{2} + 24x \\ \\ = > 12 {x}^{2} + 24x - 70x - 70 = 0 \\ \\ = > 12 {x}^{2} - 46x - 70 = 0 \\ \\ = > 2(6 {x}^{2} - 23x - 35) = 0 \\ \\ = > 6 {x}^{2} - 23x - 35 \\ \\ = > 6 {x}^{2} - 30x + 7x - 35 = 0 \\ \\ = > 6x(x - 5) + 7(x - 5) = 0 \\ \\ = > (6x + 7)(x - 5) =0$

6χ + 7 = 0 αи∂ χ - 5 = 0

χ = -7/6 αи∂ χ = 5

иєgℓє¢тιиg χ = -7/6, ωє gєт,

χ = 5.

χ + 2 = 5 + 2 = 7.

нєи¢є, тнє иυмвєяѕ αяє 5 αи∂ 7.

нσρє тнιѕ нєℓρѕ.