Step-by-step explanation:
Given expression [tex]\frac{1}{x} -\frac{2}{x^2+x}[/tex].
Let is factor second denominator [tex]x^2+x[/tex] first.
Factoring out GCF x we get
[tex]x^2+x= x(x+1)[/tex]
Rewriting the equation
[tex]\frac{1}{x} -\frac{2}{x^2+x} = \frac{1}{x} - \frac{2}{x(x+1)}[/tex]
Now, we need to find the lowest common denominator of x and x(x+1).
The lowest common denominator of x and x(x+1) is x(x+1).
So, we need to multiply first fraction by (x+1) in top and bottom to get lowest common denominator x(x+1) under first fraction.
[tex]\frac{1}{x} - \frac{2}{x(x+1)} = \frac{1(x+1)}{x(x+1)} - \frac{2}{x(x+1)}[/tex]
[tex]= \frac{x+1}{x(x+1)} - \frac{2}{x(x+1)}[/tex]
We got denominators same.
Therefore, subtracting numerators, we get
[tex]=\frac{x+1-2}{x(x+1)}[/tex]
[tex]\frac{x-1}{x(x+1)}[/tex].
Therefore, simplified form is [tex]\frac{x-1}{x(x+1)}[/tex].
