Where's the problem statement?

Notice that the answer to this problem is simultaneously $2a^2 - 14ab + 7$ and $\frac{a}{b}.$ Equating the two and doing algebraic manipulation yields

$\begin{aligned} 2a^2 - 14ab + 7 &= \dfrac{a}{b} \\ 2a^2b - 14ab^2 - a + 7b &= 0 \\ 2ab(a - 7b) - (a - 7b) &= 0 \\ (2ab - 1)(a - 7b) &= 0. \end{aligned}$

Thus, either $2ab - 1 = 0$ or $a - 7b = 0.$ The first equality gives $ab = \frac{1}{2},$ which can never be satisfied, since $a, b$ are integers. Therefore, we conclude that $a = 7b,$ so the answer to the problem is $\frac{a}{b} = \frac{7b}{b} = \boxed{7}.$

I think the question was written with inappropriate wording that would give rise to misinterpretation. There is no way to solve this hadn't I been able to interpret the question as follows:

Given that and a, b are positive integers. What is the value of $2a^{2}-14ab+7$ if the value is also equal to $\frac{a}{b}$ ?

In which case it is 7

$2a^2 - 14ab + 7 = \dfrac{a}{b} \implies 2a^2b - 14ab^2 + 7b - a = 0 \implies$

$2a^2b - a - 14ab^2 + 7b = 0 \implies a(2 ab - 1) - 7b(2ab - 1) = 0 \implies (a - 7b)(2ab - 1) = 0 \implies$

$a - 7b = 0$ or $2ab - 1 = 0$

$2ab - 1 = 0 \implies ab = \dfrac{1}{2}$ which has no solution for positive integers $a$ and $b$ .

$\therefore a = 7b \implies \dfrac{a}{b} = 7.$

The answer to the following question is an integer , which can also be expressed as $\frac{a}{b}$ . By this, $b=1$ .

The problem then reduces to: $\begin{aligned} 2a^2-14a+7=a\\ 2a^2-15a+7=0\\ (2a-1)(a-7)=0\\ a = \frac{1}{2} \text{ or } a = 7 \end{aligned}$

Since only one solution is an integer $a = \boxed{7}$ .

nice one !

$2a^2$ $-$ $14ab$ $+$ $7$ $=$ $a$ $/$ $b$

(this is the problem statement !)

cross multplying and rearranging we get

$2ab$ $(a - 7b)$ $=$ $(a - 7b)$

this statement gives $2$ possibilities,

either $ab$ $=$ $1/2$ ( the condition with this is a not equal to 7b)

$OR$

$a$ $=$ $7b$

but the former statement is not possible since $(a,b)$ are positive integers.

so $a = 7b$ so $a/b$ $=$ $7$ .