The given equations are:
\(\frac{5}x\) + 6y = 13 ……..(i)
\(\frac{3}x\) + 4y = 7 ……..(ii)
Putting \(\frac{1}x\) = u, we get:
5u + 6y = 13 …….(iii)
3u + 4y = 7 ……(iv)
On multiplying (iii) by 4 and (iv) by 6, we get:
20u + 24y = 52 ……..(v)
18u + 24y = 42 ……..(vi)
On subtracting (vi) from (v), we get:
2u = 10 ⇒ u = 5
⇒ \(\frac{1}x\) = 5 ⇒ x = \(\frac{1}5\)
On substituting x = \(\frac{1}5\) in (i), we get:
\(\frac{\frac{5}1}{3}\) + 6y = 13
25 + 6y = 13
6y = (13 – 25) = -12
y = -2
Hence, the required solution is x = \(\frac{1}5\) and y = -2.
