Solve for x and y:\(\frac{5}x\) + 6y = 13, \(\frac{3}x\) + 4y = 7

1.	Solve for x and y:\(\frac{5}x\) + 6y = 13, \(\frac{3}x\) + 4y = 7
Answer» 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.

Answer»

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.

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