Explanation
$$\Rightarrow \dfrac{3x-2}{3y+7}=\dfrac{5x-1}{5y+16}$$$$\Rightarrow 15xy-10y+48x-32=15xy+35x-3y-7$$$$\Rightarrow 13x-7y=25.........eq1$$$$\Rightarrow \dfrac{3x-15}{x-9}=\dfrac{6y-5}{2y+3}$$$$\Rightarrow 6xy-30y+9x-45=6xy-54y-5x+45$$$$\Rightarrow 14x+24y=90.......eq2$$Multiply eq1 by 24 and eq2 by 7$$\Rightarrow 312x-168y=600.......eq3$$$$\Rightarrow 98x+168y=630.......eq4$$Add eq3 and eq4$$\Rightarrow 410x=1230$$$$\Rightarrow x=3$$Put x=3 in eq 1$$\Rightarrow 13\times3-7y=25$$$$\Rightarrow -7y=-14$$$$\Rightarrow y=2$$Hence, x=3 and y=2
Let $$\dfrac {x}{y}$$ be the fraction, where $$x$$ and $$y$$ are positive integers.
$$\Rightarrow \dfrac {x+2}{y+2}=\dfrac {9}{11}$$ ..(1)$$\Rightarrow \dfrac {x+3}{y+3}=\dfrac {5}{6}$$ ..(2)
From (1) we get $$\Rightarrow 11x+22 = 9y+18$$$$\Rightarrow 11x-9y = -4$$ ...(3)
From (2) we get $$\Rightarrow 6x +18 = 5y +15$$$$\Rightarrow 6x-5y = -3$$ ...(4)
From (4) will get$$y = \dfrac{3+6x}{5}$$
Now substituting $$y$$ in (3),
$$\Rightarrow 11x-\dfrac{9\left ( 3+6x \right )}{5} = -4$$
$$\Rightarrow 55x-27-54x = -20$$
$$\Rightarrow 55x-54x = -20+27$$
$$\Rightarrow x = 7$$
Now substituting $$x$$ in (4)
$$\Rightarrow 6\times 7 -5y = -3$$
$$\Rightarrow 42-5y = -3$$
$$\Rightarrow -5y = -3 - 42$$
$$\Rightarrow -5y = -45$$
$$\Rightarrow y = 9$$
Hence, $$ x = 7$$ and $$ y = 9$$.
And fraction $$\dfrac{x}{y}=\dfrac{7}{9}$$
The given equations are
$$ \dfrac { 1 }{ x-1 } +\dfrac { 1 }{ y-2 } =2$$ and
$$ \dfrac { 6 }{ x-1 } -\dfrac { 2 }{ y-2 } =1 $$
Let $$ \dfrac { 1 }{ x-1 } =u$$ and $$ \dfrac { 1 }{ y-2 } =v$$
$$ u+v=2$$ .......(i) and
$$ 6u-2v=1$$ ......(ii)
Multiply (i) by $$6$$
$$\Rightarrow 6u+6v=12$$ ......(iii)
Subtract (ii) from (iii)
$$\Rightarrow 8v=11$$
$$\Rightarrow v=\dfrac { 11 }{ 8 } $$
Therefore $$ \dfrac { 1 }{ y-2 } =\dfrac { 11 }{ 8 } $$
$$ \Rightarrow y-2=\dfrac { 8 }{ 11 } $$
$$ \Rightarrow y=\dfrac { 30 }{ 11 } $$
Putting $$ v=\dfrac { 11 }{ 8 } $$ in (i), we get
$$ u+\dfrac { 11 }{ 8 } =2$$
$$ \Rightarrow u=\dfrac { 5 }{ 8 } $$
$$ \Rightarrow \dfrac { 1 }{ x-1 } =\dfrac { 5 }{ 8 }$$
$$ \Rightarrow x-1=\dfrac { 8 }{ 5 } $$
$$ \Rightarrow x=\dfrac { 13 }{ 5 }$$
Therefore, $$ \left( x,y \right) =\left( \dfrac { 13 }{ 5 } ,\dfrac { 30 }{ 11 } \right) $$
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