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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