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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