Two ration numbers have a sum of 17/12.

Question

17/12 The second number is formed by reducing both the numerator and denominator of the first number by one. If the denominator of the first number is one more than its  numerator, find the two fractions.  I know the answer is 3/4 and 2/3 but I do not know how to work it out to get that answer. could you help show me . thank you.

Answer 1

Let the first number be x/(x+1)

 

The second number is (x-1)/x

 

We we add two fractions x/(x+1) + (x-1)/x

 

we get a commond denominator of x(x+1)  and

 

the first numerator becomes x*x or x^2 the second numerator becomes (x+1)(x-1) or x^2 - 1

 

Therefore the sum of our numerators is 2x^2 -1

 

and that equals 17 so

 

2x^2 -1 = 17

 

2x^2 = 18

 

x^2 = 9

 

x = 3

 

our first number is 3/4  and our second number is 2/3   and when we add them together we get 17/12

Answer 2

a/(a+1) = 1st number

(a-1)/a = 2nd number

a/(a+1) + (a-1)/a = 17/12

Multiply by 12a(a+1)

12a^2 + 12(a+1)(a-1) = 17a(a+1)

12a^2 + 12(a^2-1) = 17a^2 + 17a

24a^2 - 12 = 17a^2 + 17a

7a^2 - 17a - 12 = 0

a = (- - 17 ± √((-17)^2-4(7)(-12)))/(2(7))

a = (17 ± √(289+336))/14

a = (17 ± √(625))/14

a = (17 ± 25)/14

a = -8/14, 42/14

a = -4/7, 3

1st number = a/(a+1) = 3/4
or (-4/7)/(3/7) = -4/3

2nd number = (a-1)/a = 2/3
or (-4/7 - 7/7)/(-4/7) = 11/4

check:

3/4 + 2/3 =? 17/12
(9 + 8)/12 = 17/12 √

-4/3 + 11/4 =? 17/12
(-16 + 33)/12 =? 17/12
7/12 ≠ 17/12 X

So the only solution is 3/4 and 2/3.