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Question

Solve for x and y the following system of equations:
$$\dfrac{5}{(x - y)} + \dfrac{1}{y - 2} = 2$$
$$ \dfrac{6}{x - y} - \dfrac{3}{y - 2}  = 1$$


Solution

Given pair of linear equations is 
$$\dfrac{5}{x - 1} - \dfrac{1}{y - 2} = 2...(i)$$
And $$\dfrac{6}{x - y} + \dfrac{3}{(y - 2} = 1(ii)$$
On multiplying Eq.(i) by 3  to make the coefficients equal of second term, we get the equations as
$$\dfrac{15}{x - y} + \dfrac{2}{(y - 2} = 6$$....(iii)
On substracting Eq.(iii) from Eq.(ii), we get
$$\dfrac{6}{x - 1} - \dfrac{3}{y - 2} - \dfrac{15}{x - 1} + \dfrac{3}{y - 2} = 6 + 1$$
$$ \Rightarrow \dfrac{6}{x + 1} - \dfrac{15}{x + 1} = 7$$
$$ \Rightarrow \dfrac{21}{x - 1} = 7$$
$$\Rightarrow x - 1 = 3$$
$$ \Rightarrow x = 3 + 1 $$
$$\Rightarrow x = 4$$
On putting the value of x = 4 in Eq. (Ii)we get 
$$ \dfrac{6}{4 - 1} - \dfrac{3}{y - 2} = 1$$
$$\Rightarrow 2 - \dfrac{3}{y - 2} = 1$$
$$\Rightarrow -\dfrac{3}{y - 2} = -1$$
$$\Rightarrow (y - 2) = 3$$
$$\Rightarrow y = 3 + 2$$
$$\Rightarrow y = 5$$
Hence, $$x = 4$$ and $$y = 5$$ which is the required solution 

Mathematics

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