Solution of equations $\dfrac{4a - 3}{6} + \dfrac{5b - 7}{2}$ = 18 - 5a and a + b = 5 is :

1. a = -3, b = 2

2. a = -3, b = -2

3. a = 3, b = 2

4. a = 3, b = -2

Given,

1st equation :

$\Rightarrow \dfrac{4a - 3}{6} + \dfrac{5b - 7}{2} = 18 - 5a \\[1em] \Rightarrow \dfrac{4a - 3 + 3(5b - 7)}{6} = 18 - 5a \\[1em] \Rightarrow \dfrac{4a - 3 + 15b - 21}{6} = 18 - 5a \\[1em] \Rightarrow 4a + 15b - 24 = 6(18 - 5a) \\[1em] \Rightarrow 4a + 15b - 24 = 108 - 30a \\[1em] \Rightarrow 4a + 30a + 15b = 108 + 24 \\[1em] \Rightarrow 34a + 15b = 132 \\[1em] \Rightarrow 34a + 15b - 132 = 0$

2nd equation :

⇒ a + b = 5

⇒ a + b - 5 = 0

By cross-multiplication method :

$\Rightarrow \dfrac{a}{15 \times (-5) - 1\times (-132)} = \dfrac{b}{(-132) \times 1 - (-5) \times 34} = \dfrac{1}{34 \times 1 - 1 \times 15} \\[1em] \Rightarrow \dfrac{a}{-75 + 132} = \dfrac{b}{-132 + 170} = \dfrac{1}{34 - 15} \\[1em] \Rightarrow \dfrac{a}{57} = \dfrac{b}{38} = \dfrac{1}{19} \\[1em] \Rightarrow \dfrac{a}{57} = \dfrac{1}{19} \text{ and } \dfrac{b}{38} = \dfrac{1}{19}\\[1em] \Rightarrow a = \dfrac{57}{19} \text{ and } b = \dfrac{38}{19} \\[1em] \Rightarrow a = 3 \text{ and } b = 2.$

Hence, Option 3 is the correct option.