1.

Solve the following equations by elimination method:217x + 131y = 913; 131x + 217y = 827

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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