Please help me simplify this.

$\displaystyle \frac{1}{x}+\frac{2}{x+1}$

Printable View

- Jun 1st 2009, 05:59 PMroger federeralgebraic fractions
Please help me simplify this.

$\displaystyle \frac{1}{x}+\frac{2}{x+1}$ - Jun 1st 2009, 06:11 PMartvandalay11
Well as you know, to add fractions we need a common denominator, the easiest thing to do when you have polynomials in the denominator is to create a common denominator by multiplying them together, so for this problem, our common denominator will be x(x+1)

Also remember, that we can multiply things by 1 without changing their value, so we are going to multiply by 1 in a special way

$\displaystyle \frac{1}{x}+\frac{2}{x+1}=\frac{1}{x}*\frac{x+1}{x +1}+\frac{2}{x+1}*\frac{x}{x}$ all we did was multiply each fraction by 1, but in a special form

So now lets expand, $\displaystyle \frac{x+1}{x(x+1)}+\frac{2x}{x(x+1)}$

And now we can add the two fractions, $\displaystyle =\frac{x+1+2x}{x(x+1)}=\frac{3x+1}{x(x+1)}$

I consider that as simple as possible, though some might want to multiply the denominator out, yielding $\displaystyle \frac{3x+1}{x^2+x}$