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Solve for x and y:\(\frac{1}{2(x+2y)}+\frac{5}{3(3x-2y)}=-\frac{3}2\),\(\frac{1}{4(x+2y)}-\frac{3}{5(3x-2y)}=\frac{61}{60}\) where x + 2y ≠ 0 and 3x – 2y ≠ 0 |
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Answer» The given equations are \(\frac{1}{2(x+2y)}+\frac{5}{3(3x-2y)}=-\frac{3}2\)......(i) \(\frac{1}{4(x+2y)}-\frac{3}{5(3x-2y)}=\frac{61}{60}\).......(ii) Putting 1/x+2y = u and 1/3x−2y = v, we get: 1/2 u + 5/3 v = - 3/2 ……..(iii) 5/4 u – 3/5 v = 61/60 …….(iv) On multiplying (iii) by 6 and (iv) by 20, we get: 3u + 10v = -9 …..(v) 25u – 12v = 61/3 …..(vi) On multiplying (v) by 6 and (vi) by 5, we get 18u + 60v = -54 …….(vii) 125u – 60v = 305/3 ……..(viii) On adding(vii) and (viii), we get: 143u = 305/3 – 54 = 305−162/3 = 143/3 ⇒u = 1/3 = 1/x+2y ⇒x + 2y = 3 …….(ix) On substituting u = 1/3 in (v), we get: 1 + 10v = -9 ⇒10v = -10 ⇒v = -1 ⇒ 1/3x−2y = -1 ⇒3x – 2y = -1 …….(x) On adding (ix) and (x), we get: 4x = 2 ⇒x = 1/2 On substituting x = 1/2 in (x), we get: 3/2 – 2y = -1 2y = (3/2 + 1) = 5/2 y = 5/4 Hence, the required solution is x = 1/2 and y = 5/4. |
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