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

2(x - 3) + 3(y - 5) = 0
5(x - 1) + 4(y - 4) = 0

Given,

Equations : 2(x - 3) + 3(y - 5) = 0 and 5(x - 1) + 4(y - 4) = 0

⇒ 2(x - 3) + 3(y - 5) = 0

⇒ 2x - 6 + 3y - 15 = 0

⇒ 2x + 3y - 21 = 0

⇒ 2x = 21 - 3y

⇒ x = $\dfrac{21 - 3y}{2}$ ……….(1)

⇒ 5(x - 1) + 4(y - 4) = 0

⇒ 5x - 5 + 4y - 16 = 0

⇒ 5x + 4y - 21 = 0

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

$\Rightarrow 5 \times \Big(\dfrac{21 - 3y}{2}\Big) + 4y - 21 = 0 \\[1em] \Rightarrow \dfrac{105 - 15y}{2} + 4y - 21 = 0 \\[1em] \Rightarrow \dfrac{105 - 15y + 8y - 42}{2} = 0 \\[1em] \Rightarrow 63 - 7y = 0 \\[1em] \Rightarrow 7y = 63 \\[1em] \Rightarrow y = \dfrac{63}{7} = 9.$

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

$\Rightarrow x = \dfrac{21 - 3 \times 9}{2} \\[1em] = \dfrac{21 - 27}{2} \\[1em] = -\dfrac{6}{2} \\[1em] = -3.$

Hence, x = -3 and y = 9.