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

2x - 3y - 3 = 0

$\dfrac{2x}{3} + 4y + \dfrac{1}{2} = 0$

Given,

2x - 3y - 3 = 0 ………..(i)

$\dfrac{2x}{3} + 4y + \dfrac{1}{2} = 0$ ……..(ii)

Multiplying eq. (ii) by 3 we get,

$2x + 12y + \dfrac{3}{2} = 0$ ………(iii)

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

$\Rightarrow 2x + 12y + \dfrac{3}{2} - (2x - 3y - 3) = 0 \\[1em] \Rightarrow 2x - 2x + 12y + 3y + \dfrac{3}{2} + 3 = 0 \\[1em] \Rightarrow 15y + \dfrac{9}{2} = 0 \\[1em] \Rightarrow 15y = -\dfrac{9}{2} \\[1em] \Rightarrow y = \dfrac{-9}{2 \times 15} \\[1em] \Rightarrow y = -\dfrac{3}{10}.$

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

$\Rightarrow 2x - 3y - 3 = 0 \\[1em] \Rightarrow 2x - 3 \times -\dfrac{3}{10} - 3 = 0 \\[1em] \Rightarrow 2x + \dfrac{9}{10} - 3 = 0 \\[1em] \Rightarrow 2x + \dfrac{9 - 30}{10} = 0 \\[1em] \Rightarrow 2x - \dfrac{21}{10} = 0 \\[1em] \Rightarrow x = \dfrac{21}{20}.$

Hence, $x = \dfrac{21}{20} \text{and y} = -\dfrac{3}{10}.$