Do the following pair of linear equations have no solution? Justify your answer.
$3x + y - 3 = 0$
$2x + \frac{2}{3}y = 2$

(B) No,the given pair of equations is:
$3x + y - 3 = 0$ and $2x + \frac{2}{3}y - 2 = 0$
Here,the coefficients are:
$a_1 = 3, b_1 = 1, c_1 = -3$
$a_2 = 2, b_2 = \frac{2}{3}, c_2 = -2$
Calculating the ratios:
$\frac{a_1}{a_2} = \frac{3}{2}$
$\frac{b_1}{b_2} = \frac{1}{2/3} = \frac{3}{2}$
$\frac{c_1}{c_2} = \frac{-3}{-2} = \frac{3}{2}$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{3}{2}$,the lines are coincident.
Therefore,the given pair of linear equations has infinitely many solutions,not no solution.