solve: $\frac{x - 3}{x}-\frac{3}{x + 1}+\frac{3}{x^{2}+x}=0$ $x = ?$

Explanation:

Step1: Find the common denominator

The common denominator of $x$, $x + 1$ and $x^{2}+x=x(x + 1)$ is $x(x + 1)$.

Step2: Rewrite each fraction with the common denominator

$\frac{(x - 3)(x + 1)}{x(x + 1)}-\frac{3x}{x(x + 1)}+\frac{3}{x(x + 1)}=0$.

Step3: Combine the numerators

$\frac{(x - 3)(x + 1)-3x + 3}{x(x + 1)}=0$. Expand $(x - 3)(x + 1)$: $(x - 3)(x + 1)=x^{2}+x-3x-3=x^{2}-2x-3$. So the numerator becomes $x^{2}-2x-3-3x + 3$.

Step4: Simplify the numerator

$x^{2}-2x-3-3x + 3=x^{2}-5x$. So we have $\frac{x^{2}-5x}{x(x + 1)}=0$.

Step5: Factor the numerator

$\frac{x(x - 5)}{x(x + 1)}=0$. Since $x\neq0$ (otherwise the original fractions are undefined), we can cancel out $x$ (for $x\neq0$) and get $\frac{x - 5}{x + 1}=0$.

Step6: Solve for $x$

For a fraction $\frac{a}{b}=0$ ($b\neq0$), we need $a = 0$. So $x-5=0$, which gives $x = 5$.

Answer:

$x = 5$