Since $\frac{5\left(2+x\right)}{2+x}$ and $\frac{1}{2+x}$ have the same denominator, subtract them by subtracting their numerators.
$$3+\frac{1}{\frac{5\left(2+x\right)-1}{2+x}}=2$$
Do the multiplications in $5\left(2+x\right)-1$.
$$3+\frac{1}{\frac{10+5x-1}{2+x}}=2$$
Combine like terms in $10+5x-1$.
$$3+\frac{1}{\frac{9+5x}{2+x}}=2$$
Variable $x$ cannot be equal to $-2$ since division by zero is not defined. Divide $1$ by $\frac{9+5x}{2+x}$ by multiplying $1$ by the reciprocal of $\frac{9+5x}{2+x}$.
$$3+\frac{2+x}{9+5x}=2$$
To add or subtract expressions, expand them to make their denominators the same. Multiply $3$ times $\frac{9+5x}{9+5x}$.
