Solve for x

$\dfrac{2}{3}x+10-\dfrac{x}{5}=\dfrac{36}{5}$

The $LCD$ is $15$ , multiply both sides of the equation by $15$ .

$\left(\dfrac{2}{3}x+10-\dfrac{x}{5}\right)(15)=\left(\dfrac{36}{5}\right)(15)$

$10x+150-3x=108$

Combine like terms.

$7x+150=108$

Transpose $150$ to the other side of the equation.

$7x=108-150$

Combine like terms.

$7x=-42$

Divide both sides $7$ .

$\boxed{x=-6}$

Verify the solution by substitution.

$10(-6)+150-3(-6)=108$

$-60+150+18=108$

$\boxed{108=108}$ $\text{The statement is true. Therefore, -6 is the solution.}$

here it is:

$\frac {2 x }{ 3}$ + 10 - $\frac {x}{5}$ = $\frac {36}{5}$

or, $\frac {2 x }{ 3}$ - $\frac {x}{5}$ = $\frac {36}{5}$ - 10

or, $\frac {10 x -3 x }{15}$ = $-210$

or, 35x =-210

or, x= -6