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

2x + 3y = 8
2x = 2 + 3y

Given,

Equations : 2x + 3y = 8 and 2x = 2 + 3y

⇒ 2x + 3y = 8

⇒ 2x = 8 - 3y

⇒ x = $\dfrac{8 - 3y}{2}$ …………(1)

Substituting value of x from equation (1) in 2x = 2 + 3y, we get :

$\Rightarrow 2 \times \Big(\dfrac{8 - 3y}{2}\Big) = 2 + 3y \\[1em] \Rightarrow 8 - 3y = 2 + 3y \\[1em] \Rightarrow 3y + 3y = 8 - 2 \\[1em] \Rightarrow 6y = 6 \\[1em] \Rightarrow y = \dfrac{6}{6} = 1.$

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

$\Rightarrow x = \dfrac{8 - 3 \times 1}{2} \\[1em] = \dfrac{8 - 3}{2} \\[1em] = \dfrac{5}{2} = 2.5$

Hence, x = 2.5 and y = 1.