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

0.2x + 0.1y = 25
2(x - 2) - 1.6y = 116

Given,

Equations : 0.2x + 0.1y = 25 and 2(x - 2) - 1.6y = 116

⇒ 0.2x + 0.1y = 25

⇒ 0.2x = 25 - 0.1y

⇒ x = $\dfrac{25 - 0.1y}{0.2}$ ……….(1)

$\Rightarrow 2(x - 2) - 1.6y = 116 \\[1em] \Rightarrow 2x - 4 - 1.6y = 116 \\[1em] \Rightarrow 2x - 1.6y = 116 + 4 \\[1em] \Rightarrow 2x - 1.6y = 120 \\[1em]$

Substituting value of x from equation (1) in above equation, we get :

$\Rightarrow 2 \times \Big(\dfrac{25 - 0.1y}{0.2}\Big) - 1.6y = 120 \\[1em] \Rightarrow \dfrac{25 - 0.1y}{0.1} - 1.6y = 120 \\[1em] \Rightarrow \dfrac{25 - 0.1y - 0.16y}{0.1} = 120 \\[1em] \Rightarrow 25 - 0.26y = 120 \times 0.1 \\[1em] \Rightarrow 25 - 0.26y = 12 \\[1em] \Rightarrow 0.26y = 25 - 12 \\[1em] \Rightarrow 0.26y = 13 \\[1em] \Rightarrow y = \dfrac{13}{0.26} = \dfrac{1300}{26} = 50.$

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

$\Rightarrow x = \dfrac{25 - 0.1 \times 50}{0.2} \\[1em] \Rightarrow x = \dfrac{25 - 5}{0.2} \\[1em] \Rightarrow x = \dfrac{20}{0.2} = 100.$

Hence, x = 100 and y = 50.