Solve the following pairs of linear (simultaneously) equations using method of elimination by substitution:

1.5x + 0.1y = 6.2
3x - 0.4y = 11.2

Given,

Equations : 1.5x + 0.1y = 6.2 and 3x - 0.4y = 11.2

⇒ 1.5x + 0.1y = 6.2

⇒ 1.5x = 6.2 - 0.1y

⇒ x = $\dfrac{6.2 - 0.1y}{1.5}$ ……..(1)

Substituting value of x from equation (1) in 3x - 0.4y = 11.2, we get :

$\Rightarrow 3 \times \Big(\dfrac{6.2 - 0.1y}{1.5}\Big) - 0.4y = 11.2 \\[1em] \Rightarrow \dfrac{6.2 - 0.1y}{0.5} - 0.4y = 11.2 \\[1em] \Rightarrow \dfrac{6.2 - 0.1y - 0.2y}{0.5} = 11.2 \\[1em] \Rightarrow 6.2 - 0.3y = 11.2 \times 0.5 \\[1em] \Rightarrow 6.2 - 0.3y = 5.6 \\[1em] \Rightarrow 0.3y = 6.2 - 5.6 \\[1em] \Rightarrow 0.3y = 0.6 \\[1em] \Rightarrow y = \dfrac{0.6}{0.3} = 2.$

Substituting value of y in equation (1), we get :

$\Rightarrow x = \dfrac{6.2 - 0.1 \times 2}{1.5} \\[1em] = \dfrac{6.2 - 0.2}{1.5} \\[1em] = \dfrac{6}{1.5} \\[1em] = 4.$

Hence, x = 4 and y = 2.