Solve the following equations by elimination method:29x – 23y = 110;

1.	Solve the following equations by elimination method:29x – 23y = 110; 23x – 29y = 98
Answer» 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.

Solve the following equations by elimination method:29x – 23y = 110; 23x – 29y = 98

Answer»

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.