Given pair of linear equations is
29x – 23y = 110 …(i)
And 23x – 29y = 98 …(ii)
On multiplying Eq. (i) by 23 and Eq. (ii) by 29 to make the coefficients of x equal, we get the equation as
667x – 529y = 2530 …(iii)
667x – 841y = 2842 …(iv)
On subtracting Eq. (iii) from Eq. (iv), we get
⇒ 667x – 841y – 667x + 529y = 2842 – 2530
⇒ – 312y = 312
⇒ y = – 1
On putting y = 2 in Eq. (ii), we get
⇒ 29x – 23( – 1) = 110 ⇒ 29x + 23 = 110
⇒ 29x = 110 – 23
⇒ 29x = 87
⇒ x = 3
Hence, x = 3 and y = – 1 , which is the required solution.
