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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