Solve the following system of equations by elimination method:

0.4x – 1.5y = 6.5

0.3x + 0.2y = 0.9 (2000, 2004, 2005)

Given pair of linear equations is

0.4x – 1.5y = 6.5 …(i)

And 0.3x + 0.2y = 0.9 …(ii)

On multiplying Eq. (i) by 3 and Eq. (ii) by 4 to make the coefficients of x equal, we get the equation as

1.2x – 4.5y = 19.5 …(iii)

1.2x + 0.8y = 3.6 …(iv)

On subtracting Eq. (iii) from Eq. (iv), we get

1.2x + 0.8y – 1.2x + 4.5y = 3.6 – 19.5

⇒ 5.3y = – 15.9

⇒ y = – 3

On putting y = – 3 in Eq. (ii), we get

0.3x + 0.2y = 0.9

⇒ 0.3x + 0.2( – 3) = 0.9

⇒ 0.3x – 0.6 = 0.9

⇒ 0.3x = 1.5

⇒ x = 1.5/0.3

⇒ x = 5

Hence, x = 5 and y = – 3 , which is the required solution.