Find the number whose double is 45 greater than its half.

Let the number be x.

Double of the number is 45 greater than its half.

$⇒ 2x = 45 + \dfrac{1}{2}x\\[1em] ⇒ 2x - \dfrac{1}{2}x = 45 \\[1em] ⇒ \dfrac{2x}{1} - \dfrac{x}{2} = 45\\[1em]$

Since L.C.M. of denominators 1 and 2 = 2, multiply each term with 2 to get:

$⇒ \dfrac{2x \times 2}{1} - \dfrac{x \times 2}{2} = 45\times 2$

⇒ 2x $\times$ 2 - x = 90

⇒ 4x - x = 90

⇒ 3x = 90

⇒ x = $\dfrac{90}{3}$

⇒ x = 30

Hence, the number is 30.