Solve for x and y:2x + 5y = \(\frac{8}3\),3x – 2y = \(\frac{5}6\)

1.	Solve for x and y:2x + 5y = \(\frac{8}3\),3x – 2y = \(\frac{5}6\)
Answer» The given equations are: 2x – 5y = \(\frac{8}3\).....(i) 3x – 2y = \(\frac{5}6\) On multiplying (i) by 2 and (ii) by 5, we get: 4x - 10y = \(\frac{16}3\).....(iii) 15x – 10y = \(\frac{25}6\).....(iv) On adding (iii) and (iv), we get: 19x = \(\frac{57}6\) ⇒ x = \(\frac{57}{6\times19}\) = \(\frac{3}6=\frac{1}2\) On substituting x = \(\frac{1}2\) in (i), we get: 2 x \(\frac{1}2\) + 5y = \(\frac{8}3\) ⇒ 5y = \((\frac{8}3-1)\) = \(\frac{5}3\) ⇒ y = \(\frac{5}{3\times5}\) = \(\frac{1}3\) Hence, the solution is x = \(\frac{1}2\) and y = \(\frac{1}3\).

Answer»

The given equations are:

2x – 5y = \(\frac{8}3\).....(i)

3x – 2y = \(\frac{5}6\)

On multiplying (i) by 2 and (ii) by 5, we get:

4x - 10y = \(\frac{16}3\).....(iii)

15x – 10y = \(\frac{25}6\).....(iv)

On adding (iii) and (iv), we get:

19x = \(\frac{57}6\)

⇒ x = \(\frac{57}{6\times19}\) = \(\frac{3}6=\frac{1}2\)

On substituting x = \(\frac{1}2\) in (i), we get:

2 x \(\frac{1}2\) + 5y = \(\frac{8}3\)

⇒ 5y = \((\frac{8}3-1)\) = \(\frac{5}3\)

⇒ y = \(\frac{5}{3\times5}\) = \(\frac{1}3\)

Hence, the solution is x = \(\frac{1}2\) and y = \(\frac{1}3\).

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