Answers

2016-04-09T01:12:23+05:30
Let the fraction be x/y

Given that
x+2/y =1/2 -----> equ i
x/y-1 =1/3 -----> equ ii

From i and ii
2(x+2) =1(y) -------> equ i
3(x) =1(y-1) -------> equ ii

=> 2x+4=y
3x=y-1

Subtract i and ii. we get

2x - y = -4
3x - y = -1
(-) (+) = (+)
______________
-x = -3
_____________

=> x=3
subtitute x=3 in equ i. we get

2(3+2)=y
=> 2(5) =y
=> y=10

Therefore the value of x and y are 3 and 10 respectively.
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2016-04-09T08:41:02+05:30
Given,
 \frac{x+2}{y}  \frac{1}{2} ...............(1)
 \frac{x}{y-1}  \frac{1}{3} .................(2)
(1) ⇒ 2x+4 = y
      ⇒2x-y+4 = 0
(2) ⇒ 3x = y-1
      ⇒ 3x-y+1 = 0

solving the equations
3x-y+1 = 0
2x-y+4 = 0
---------------
x-3 = 0
x = 3

substitute x value in any of the above equation,
6-y+4 = 0
y = 10
0