Question 126
$$\frac{4}{x - 3} + \frac{6}{y - 4} = 5, \frac{5}{x - 3} - \frac{3}{y - 4} = 1 \Rightarrow x + y =$$
Solution
we have given that
$$\frac{4}{x - 3} + \frac{6}{y - 4} = 5$$ equation 1
$$\frac{5}{x - 3} - \frac{3}{y - 4} = 1$$ equation 2
now 2$$\times{equation}$$ - equation 1
2{$$\frac{5}{x - 3} - \frac{3}{y - 4} = 1$$} -
$$\frac{4}{x - 3} + \frac{6}{y - 4} = 5$$
then we get x= 5 put the value of x in equation 1
$$\frac{4}{5 - 3} + \frac{6}{y - 4} = 5$$ we get y = 6
then x + y = 11 answer
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