Given equations are
0.5x + 0.8y = 3.4 …(i)
0.6x – 0.3y = 0.3 …(ii)
From eqn (ii), 2x – y = 1
y = 2x – 1 …(iii)
On substituting y = 0.2x – 1 in eqn (i), we get
⇒ 0.5x + 0.8(2x – 1) = 3.4
⇒ 0.5x + 1.6x – 0.8 = 3.4
⇒ 2.1x = 3.4 + 0.8
⇒ 2.1x = 4.2
⇒ x = 4.2/2.1 = 2
Now, on putting x = 2 in eqn (iii), we get
⇒ y = 2(2) – 1
⇒ y = 4 – 1
⇒ y = 3
Thus, x = 2 and y = 3 is the required solution.