Increasing the Reciprocal

$\frac{x}{x - 2} = \frac{x - 2}{x} + \frac{16}{15} \\\\ 15x \cdot x = 15(x - 2) \cdot (x - 2) + x(x - 2) \cdot 16 \\\\ 15x^2 = 15x^2 - 60x + 60 + 16x^2 - 32x \\\\ 16x^2 - 92x + 60 = 0 \;\; \div (4 \\\\ 4x^2 - 23x + 15 = 0 \\\\ \left ( x - 5 \right )\left ( x - \frac{3}{4} \right ) = 0 \\\\ x - 5 = 0 \\\\ \boxed{x = 5}$

Daí,

$\frac{x - 2}{x} = \frac{a}{b} \\\\ \frac{5 - 2}{5} = \frac{a}{b} \\\\ \frac{3}{5} = \frac{a}{b} \\\\ a + b = 3 + 5 \\\\ \boxed{\boxed{a + b = 8}}$

$\frac{b}{a}-\frac{a}{b}=\frac{16}{15}$

$\frac{b^{2}-a^{2}}{ab}=\frac{16}{15}$

$\frac{(b+a)(b-a)}{ab}=\frac{16}{15}$

$ab=15$ a and b are either 1 and 15, or 3 and 5. Plugging 3 and 5 gets us this:

$(5+3)(5-3)=16$

$8(2)=16$

$16=16$

so a and b are 5 and 3. $5+3=\boxed{8}$

Well, let a=value of numerator and let b=value of denominator. We know that the value of the numerator is 2 less than the value of the denominator so we can say a=b-2. So our original fraction which was a/b is now b-2/b. Next, this fraction has been interchanged which means that the numerator and denominator have been switched so our new fraction is b/b-2. It says that the original fraction has been increased by 1 1/15 which can also be written as 16/15. So our equation is b-2/b + 16/15 = b/b-2. By solving this equation we get the quadratic, -4b^2+23b-15. Using the quadratic formula, we get 2 roots which are 5 and 3/4. Since our answer has to be an integer, b is 5. Since the denominator is 5, the numerator is b-2 or 5-2 which is 3. Finally a+b is (b-2)+b; (5-2)+5; 3+5= 8 .

given that a=b-2, then you can make up an equation b/a= a/b +16/15 by the given statement above. then substitute the value of a, then it will be b/(b-2)=(b-2)/b + 16/15. then simplified it and you would get a value of 3/4 and 5. But since the value must be greater than 2 to come up with positive answer. Therefore the value is 5 and the value of a is 3. By just adding it you would get 8.

Let the den. be x. Therefore, num. is x-2. therefore, x-2/x - x/x-2 = 16/15 solving this eqn. , we get: x=-3,fraction = 5/3 therefore, 5+3 = 8

a/b =b/a-16/15 after solving 15a^2+16ab-15b^2=o given a=b-2 so after putting this value we got b=5 , a=3

a/b is positive fraction ,the numerator is 2 less than the denominator[a+2=b] lets a/b=x ,b/a=1/x ,If the numerator and denominator are interchanged, the fraction is increased by 16/15 [1/x=x+16/15] solve x=3/5 or x=-5/3[not suitable]because fraction is positive x=3/5 hence a=3b/5 and a+2=b solve that b=5,a=3 a+b=5+3=8

Let the numerator = a and denominator = b

Our fraction is a/b

The numerator is 2 less than denominator

So, a = b- 2

Our new fraction is b - 2/ b. Let's say the result is x.

Then

(b - 2)/ b = x -------------------------------------------------- ( i )

The next condition says that if the numerator and denominator are interchanged the result is 16 / 15 more than previous equation.

b / (b - 2) = x + ( 16/15 ) ---------------------------------------------- ( ii )

Substituting the value of x from equation ( i ) in equation ( ii ) we get,

b / (b - 2) = (b - 2)/ b + ( 16/15 )

or, (b / (b - 2)) - ((b - 2)/ b) = ( 16/15 )

Taking L.C.M. of the denominators as b ( b -2 )

((b ^ 2) - ((b - 2) ^ 2))/ (b ( b -2 )) = ( 16/15 )

Using the formula of ((a - b) ^ 2)) for ((b - 2) ^ 2)). We get

(b ^ 2) - (b ^ 2) + 4b + 4 / ((b^2) - 2b) = ( 16/15 )

4b + 4 / ((b^2) - 2b) = ( 16/15 )

By cross multiplication,

15 ( 4b - 4 ) = 16 (( b^2) - 2b)

60b - 60 = 16( b^2) - 32b

60b + 32b = 16( b^2) + 60

92b = 16( b^2) + 60

16( b^2) - 92b + 60 = 0

16( b^2) - 80b - 12b + 60 = 0

16b ( b - 5 ) - 12 ( b - 5) = 0

(16b - 12) (b - 5) = 0

Either, 16b - 12 = 0 _---------------------------------------( iii )

or, b - 5 = 0 -----------------------------------------------( iv )

From equation ( iii )

16b - 12 = 0

16b = 12

b = 12 / 16

b = 3 / 4

If we substitute value of b in the fraction then our fraction comes as -5/3 which is not our result. So, this value of b will not work.

Let's see another equation

From equation ( iv )

b - 5 = 0

b = 5

So, our denominator (b) is 5 and our numerator (a) is 3.

a + b = 8

denominator is 15 = 3 x 5 = (5 - 2) x 5 so a + b = 3 + 5 = 8

\frac{b}{b-2}=\frac{b-2}{b}+\frac{16}{15} \frac{b}{b-2}=\frac{15b-30+16b}{15b} Multiplicando cruzado temos 15b²=15b²-30b+16b²-30b+60-32b 16b²-92b+60=0-> Encontramos essa equação cujo conjunto solução é 5 e 3/4(Não convém). Logo temos que se b=5, e a =b-2 ,logo,a=3 e \boxed{a+b=8}

(x-2)/x=a/b…(1) x/(x-2)= a/b+16/15…(2) Substitute (1)—>(2) x/(x-2)=(x-2)/x+ 16/15 x/(x-2)= (31x-30)/15x 15x^2=31x^2-92x+60 0=16x^2-92x+60 make it simple, we will get : 0= 4x^2-23x+15 0=(4x-3)(x-5) x=3/4 or x=5 We need a positive solution, so we choose x=5 Substitute x=5 to the (1) we will find the fraction. That is 3/5 So a+b=3+5=8