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Solve the following equations by elimination method:217x + 131y = 913; 131x + 217y = 827 |
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Answer» Given pair of linear equations is 217x + 131y = 913 …(i) And 131x + 217y = 827 …(ii) On multiplying Eq. (i) by 131 and Eq. (ii) by 217 to make the coefficients of x equal, we get the equation as 28427x + 17161y = 119603 …(iii) 28427x + 47089y = 179459 …(iv) On subtracting Eq. (iii) from Eq. (iv), we get ⇒ 28427x + 47089y – 28427x – 17161y = 179459 – 119603 ⇒ 47089y – 17161y = 179459 – 119603 ⇒ 29928y = 59856 ⇒ y = 59856/29928 ⇒ y = 2 On putting y = 2 in Eq. (ii), we get ⇒ 131x + 217(2) = 827 ⇒ 131x + 434 = 827 ⇒ 131x = 393 ⇒ x = 393/131 ⇒ x = 3 Hence, x = 3 and y = 2 , which is the required solution. |
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