Solve the following systems of simultaneous linear equations by the elimination method

$\dfrac{3}{4}x - \dfrac{2}{3}y = 1$

$\dfrac{3}{8}x - \dfrac{1}{6}y = 1$

Given,

$\dfrac{3}{4}x - \dfrac{2}{3}y = 1$ ……..(i)

$\dfrac{3}{8}x - \dfrac{1}{6}y = 1$ ………(ii)

Multiplying eq. (ii) by 2 we get,

$\dfrac{3}{4}x - \dfrac{1}{3}y = 2$ ……..(iii)

Subtracting eq. (i) from (iii) we get,

$\Rightarrow \dfrac{3}{4}x - \dfrac{1}{3}y - \Big(\dfrac{3}{4}x - \dfrac{2}{3}y\Big) = 2 - 1 \\[1em] \Rightarrow \dfrac{3}{4}x - \dfrac{3}{4}x - \dfrac{1}{3}y + \dfrac{2}{3}y = 1 \\[1em] \Rightarrow \dfrac{1}{3}y = 1 \\[1em] \Rightarrow y = 3.$

Substituting value of y in eq. (i) we get,

$\Rightarrow \dfrac{3}{4}x - \dfrac{2}{3} \times 3 = 1 \\[1em] \Rightarrow \dfrac{3}{4}x - 2 = 1 \\[1em] \Rightarrow \dfrac{3}{4}x = 3 \\[1em] \Rightarrow x = \dfrac{3 \times 4}{3} \\[1em] \Rightarrow x = 4.$

Hence, x = 4 and y = 3.