Standard Algebra with Unknown Integers

$\begin{aligned} 2a + 3b & = c\\ 2a + 3b - c & = 0 \longrightarrow \boxed{1}\\ \\ a + b + c & = 42 \longrightarrow \boxed{2}\\ \\ c - a & = 19\\ -a + c & = 19 \longrightarrow \boxed{3}\\ \\ \text{Multiplying} \boxed{2} \text{by 3, we get}:-\\ 3a + 3b + 3c & = 126 \longrightarrow \boxed{4}\\ \\ \text{Subtracting} \boxed{1} \text{from} \boxed{4} \text{we get}:-\\ a + 4c & = 126 \longrightarrow \boxed{5}\\ \\ \text{Adding} \boxed{3} \text{and} \boxed{5} \text{we get}:-\\ 5c & = 145\\ c & = \color{#3D99F6}{29}\\ \\ \text{Substituting c = 29 in} \boxed{3} \text{we get}:-\\ -a + 29 & = 19\\ a & = \color{#D61F06}{10}\\ \\ \text{Substituting c=29 and a=10 in} \boxed{2} \text{we get}:-\\ 10 + b + 29 & = 42\\ b & = \color{#20A900}{3}\\ \\ \implies \dfrac{c + 1}{ab} & = \dfrac{\color{#3D99F6}{29} + 1}{\color{#D61F06}{10} × \color{#20A900}{3}}\\ & = \dfrac{30}{30}\\ & = \color{#69047E}{\boxed{1}} \end{aligned}$
$\boxed{\text{Q.E.D}}$

$2a+3b=c \implies$ Eq.(1)

$a+b+c = 42 \implies$ Eq.(2)

$c-a=19 \implies c=a+19 \implies$ Eq.(3)

Substitute Eq.(3) into Eq.(1) and Eq.(2):

$2a + 3b = a+19 \implies a+3b=19 \implies$ Eq.(4)

$a+b+a+19 = 42 \implies b=23-2a \implies$ Eq.(5)

Substitute Eq.(5) into Eq.(4):

$a+3(23-2a)=19\\ a+69-6a=19\\ 5a=50\\ a=10$

Use this to find $b$ and $c$ :

$b=23-2(10) = 3\\ c=10+19=29$

$\dfrac{c+1}{ab} = \dfrac{29+1}{10 \times 3} = \dfrac{30}{30} = \boxed{1}$

Solving Equation 3) for $c$ yields $c=19 +a$ while solving Equation 1) for $b$ yields $b =\frac{c - 2a}{3}=\frac{19 + a - 2a}{3}=\frac{19 - a}{3}$

Now, plugging all this into equation 2) yields $a+\frac{19 - a}{3}+19+a =42$ .

$\frac{19 - a}{3}+2a=42-19 \\ \frac{19 - a + 6a}{3}=23 \\ 19 + 5a = 23\times3 = 69 \\ 5a = 69 - 19 = 50 \\ a=\frac{50}{5}=10$ .

Knowing $a$ , it is possible to solve for both $b$ and $c$ .

$b=\frac{19 - a}{3}=\frac{19 - 10}{3}=\frac{9}{3}= 3 \implies b = 3. \\ c = 19 +a = 19 + 10 = 29$ .

$a = 10; b = 3; c = 29$

$\therefore \frac{c+1}{ab}=\frac{29+1}{10 \times 3}=\frac{30}{30}= \boxed{1}$ .