A Tricky Fractional Equation

Replace $*$ with $x$ , then:

$\dfrac {2}{x} - \dfrac {x}{5} = \dfrac {1}{15}\quad \Rightarrow \dfrac {10-x^2}{5x} = \dfrac {1}{15} \quad \Rightarrow 30-3x^2 = x$

$\Rightarrow 3x^2+x-30 = 0 \quad \Rightarrow (3x+10)(x-3) = 0$

For a whole number $*$ , $\quad \Rightarrow * = x =\boxed{3}$

$\frac{2}{3}-\frac{3}{5}=\frac{1}{15}$

$*=3$

Sometimes is easier substitute all the values than solve a cuadratic equation xD

Also $3\times 5$ is the only product equal $15$

$\frac{2}{X}-\frac{X}{5}=\frac{1}{3\times5}$ We can deduce from here that $X$ must be a multiple of $3$ . By verification, we can easily calculate that the number $3$ itself is a solution.

first we can simplify the equation to get a quadratic equation from which we can say x will be positive and it will be equal to 3 . so x=3

2/x-x/5=1/15

=15x^2+5x-150=0

=3x^2+x-30=0

=3x^2-9x+10x-30=0

=3x(x-3)+10(x-3)=0

=(x-3)(3x+10)=0

= x-3=0 or 3x+10=0

=x=3 or x=-3/10

but as x is also denominator x cannot be 0

therefore x=0

simply substitute the options and get the desired answer