Answer :
1/X+ 1/X+2= 28/195
Common denominator. 195x(X+5)
195(X+2)+195x= 28x(X+2)
28x-^2+56x-390c-390=0
Do quadratic formula
-(-334) +- sq rt (-334)^2-4(28)(-390). /56
334+ 394 /56
728/56= 13
Answer 1/13 , 1/15 />
1/13+1/15= 28/195
15/195+13195= 28/195
Common denominator. 195x(X+5)
195(X+2)+195x= 28x(X+2)
28x-^2+56x-390c-390=0
Do quadratic formula
-(-334) +- sq rt (-334)^2-4(28)(-390). /56
334+ 394 /56
728/56= 13
Answer 1/13 , 1/15 />
1/13+1/15= 28/195
15/195+13195= 28/195
Building and solving an equation using the reciprocal concept, it is found that the integers are 13 and 15.
- The reciprocal of an integer x is given by [tex]\frac{1}{x}[/tex].
- In this problem, we have two consecutive odd integers, which are [tex]x[/tex] and [tex]x + 2[/tex], and their reciprocals are: [tex]\frac{1}{x}[/tex] and [tex]\frac{1}{x + 2}[/tex]
The sum of the reciprocals is [tex]\frac{28}{195}[/tex], thus:
[tex]\frac{1}{x} + \frac{1}{x + 2} = \frac{28}{195}[/tex]
[tex]\frac{x + 2 + x}{x(x + 2)} = \frac{28}{195}[/tex]
[tex]\frac{2x + 2}{x(x + 2)} = \frac{28}{195}[/tex]
[tex]28x^2 + 56x = 390x + 390[/tex]
[tex]28x^2 - 334x - 390 = 0[/tex]
Which is a quadratic equation with coefficients [tex]a = 28, b = -334, c = -390[/tex].
The solutions to this equation are given by:
[tex]\Delta = (-334)^2 - 4(28)(-390) = 155236[/tex]
[tex]x_{1} = \frac{334 + \sqrt{155236}}{2(28)} = 13[/tex]
[tex]x_{2} = \frac{334 - \sqrt{155236}}{2(28)} = -1.07[/tex]
The numbers are integers, so [tex]x = 13[/tex], [tex]x + 2 = 13 + 2 = 15[/tex], thus the two integers are 13 and 15.
A similar problem is given at https://brainly.com/question/24764843