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

$2x - \dfrac{3y}{4} = 3$

5x - 2y - 7 = 0

Given,

$2x - \dfrac{3y}{4} = 3$ …….(i)

5x - 2y - 7 = 0 …….(ii)

Multiplying eqn. (i) by 4 we get,

$\Rightarrow 4\Big(2x - \dfrac{3y}{4}\Big) = 3 \times 4 \\[1em] \Rightarrow 8x - 3y = 12 \\[1em] \Rightarrow 8x = 12 + 3y \\[1em] \Rightarrow x = \dfrac{12 + 3y}{8} ……(iii)$

Putting value of x from eqn. (iii) in eqn. (ii),

$\Rightarrow 5\Big(\dfrac{12 + 3y}{8}\Big) - 2y - 7 = 0 \\[1em] \Rightarrow \dfrac{60 + 15y}{8} - 2y - 7 = 0 \\[1em] \Rightarrow \dfrac{60 + 15y - 16y - 56}{8} = 0 \\[1em] \Rightarrow 4 - y = 0 \\[1em] \Rightarrow y = 4.$

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

$\Rightarrow x = \dfrac{12 + 3y}{8} \\[1em] = \dfrac{12 + 3(4)}{8} \\[1em] = \dfrac{12 + 12}{8} \\[1em] = \dfrac{24}{8} \\[1em] = 3.$

Hence, x = 3 and y = 4.