Solve the following system of equations:4/x + 3y = 14;3/x – 4y = 23

1.	Solve the following system of equations:4/x + 3y = 14;3/x – 4y = 23
Answer» Taking 1/x = u, the given equation becomes 4u + 3y = 14…………(i) 3u – 4y = 23…………(ii) Adding (i) and (ii), we get 4u + 3y + 3u – 4y = 14 + 23 ⇒ 7u – y = 37 ⇒ y = 7u – 37…………(iii) Using (iii) in (i), 4u + 3(7u – 37) = 14 ⇒ 4u + 21u – 111 = 14 ⇒ 25u = 125 ⇒ u = 5 ⇒ x = 1/u = \(\frac{1}{5}\) Putting u= 5 in (iii), we find y y = 7(5) – 37 ⇒ y = -2 Hence, the solution for the given system of equations is x = \(\frac{1}{5}\) and y = -2.

Answer»

Taking 1/x = u, the given equation becomes

4u + 3y = 14…………(i)

3u – 4y = 23…………(ii)

Adding (i) and (ii), we get

4u + 3y + 3u – 4y = 14 + 23

⇒ 7u – y = 37

⇒ y = 7u – 37…………(iii)

Using (iii) in (i),

4u + 3(7u – 37) = 14

⇒ 4u + 21u – 111 = 14

⇒ 25u = 125

⇒ u = 5

⇒ x = 1/u = \(\frac{1}{5}\)

Putting u= 5 in (iii), we find y

y = 7(5) – 37

⇒ y = -2

Hence, the solution for the given system of equations is x = \(\frac{1}{5}\) and y = -2.

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