# How do you simplify and find the restrictions for (x^2+x)/(x^2+2x)?

Jul 14, 2017

(x+1)/(x+2); x!=-2, 0

#### Explanation:

$\frac{{x}^{2} + x}{{x}^{2} + 2 x}$

$= \frac{\cancel{x} \left(x + 1\right)}{\cancel{x} \left(x + 2\right)}$ $\to$ factor and cancel

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

The restrictions consist of $x$-values that make the denominator zero.

Set the original expression's denominator equal to $0$:

${x}^{2} + 2 x \ne 0$
$x \left(x + 2\right) \ne 0$
$x \ne 0$ and $x + 2 \ne 0 \implies x \ne - 2$

Set the simplified expression's denominator equal to $0$:
$x + 2 \ne 0$
$x \ne - 2$
This is the same answer as above, but sometimes there are additional restricted values that result from the simplified expression.

(x+1)/(x+2); x!=-2, 0

Jul 14, 2017

$\frac{x + 1}{x + 2}$ [$x \ne 0$ and $x \ne - 2$]

#### Explanation:

Expression $= \frac{{x}^{2} + x}{{x}^{2} + 2 x}$

$= \frac{x \left(x + 1\right)}{x \left(x + 2\right)}$

If $x \ne 0$

Expression $= \frac{x + 1}{x + 2}$

Hence, Expression is defined $\forall x \in \mathbb{R} \ne - 2$

$\therefore$Expression $= \frac{x + 1}{x + 2}$ [$x \ne 0$ and $x \ne - 2$]