The number of integral solutions of the equation $7\left(y+\frac{1}{y}\right)-2\left(y^2+\frac{1}{y^2}\right)=9 \text { is }$

We have:

$7\left(y+\dfrac{1}{y}\right)-2\left(y^2+\dfrac{1}{y^2}\right)=9$

Note that $y \neq 0$ since we have $\dfrac{1}{y}$ terms.

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Let $t = y + \dfrac{1}{y}$

Squaring both sides:

$t^2 = y^2 + 2 + \dfrac{1}{y^2}$

So,

$y^2 + \dfrac{1}{y^2} = t^2 - 2$

This is a classic substitution — whenever you see $y + \dfrac{1}{y}$ and $y^2 + \dfrac{1}{y^2}$ together, always try this.

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Substituting into the original equation:

$7(t) - 2(t^2 - 2) = 9$

$7t - 2t^2 + 4 = 9$

$2t^2 - 7t + 5 = 0$

$(2t - 5)(t - 1) = 0$

$t = \dfrac{5}{2} \quad \text{or} \quad t = 1$

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Case 1: $t = \dfrac{5}{2}$

$y + \dfrac{1}{y} = \dfrac{5}{2}$

Multiplying both sides by $y$:

$2y^2 - 5y + 2 = 0$

$(2y - 1)(y - 2) = 0$

$y = \dfrac{1}{2} \quad \text{or} \quad y = 2$

Only $y = 2$ is an integer.

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Case 2: $t = 1$

$y + \dfrac{1}{y} = 1$

Multiplying both sides by $y$:

$y^2 - y + 1 = 0$

Discriminant $= (-1)^2 - 4(1)(1) = 1 - 4 = -3 < 0$

No real solutions exist for this case.

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Verification for $y = 2$:

$7\left(2 + \dfrac{1}{2}\right) - 2\left(4 + \dfrac{1}{4}\right)$

$= 7\left(\dfrac{5}{2}\right) - 2\left(\dfrac{17}{4}\right)$

$= \dfrac{35}{2} - \dfrac{17}{2}$

$= \dfrac{18}{2} = 9 \checkmark$

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The number of integral solutions is $\boxed{1}$.