Solve the following system of simultaneous linear equations by the substitution method:

2x + 3y = 23

5x - 20 = 8y

Given,

2x + 3y = 23 ……(i)

5x - 20 = 8y ……(ii)

Solving (i) we get,

⟹ 2x + 3y = 23

⟹ 2x = 23 - 3y

⟹ x = $\dfrac{23 - 3y}{2}$ …..(iii)

Substituting value of x from eqn. (iii) in eqn. (ii) we get,

$\Rightarrow 5\Big(\dfrac{23 - 3y}{2}\Big) - 20 = 8y \\[1em] \Rightarrow \dfrac{115 - 15y}{2} - 20 = 8y \\[1em] \Rightarrow \dfrac{115 - 15y - 40}{2} = 8y \\[1em] \Rightarrow 115 - 15y - 40 = 16y \\[1em] \Rightarrow 75 - 15y = 16y \\[1em] \Rightarrow 31y = 75 \\[1em] \Rightarrow y = \dfrac{75}{31} = 2\dfrac{13}{31}.$

Substituting value of y in eqn.(iii) we get,

$\Rightarrow x = \dfrac{23 - 3y}{2} \\[1em] = \dfrac{23 - 3 \times \dfrac{75}{31}}{2} \\[1em] = \dfrac{\dfrac{23 \times 31 - 3 \times 75}{31}}{2} \\[1em] = \dfrac{713 - 225}{31 \times 2} \\[1em] = \dfrac{488}{62} \\[1em] = \dfrac{244}{31} = 7\dfrac{27}{31}.$

Hence, x = $7\dfrac{27}{31}\text{ and y =} 2\dfrac{13}{31}$.