Answer
$\dfrac{3(x + 1)}{2(x + 3)}$
Restrictions: $x \ne -3, -1, 1$
Work Step by Step
First, factor all expressions in the original exercise:
$\dfrac{5(x + 3)}{10(x - 1)} \div \dfrac{(x + 3)(x + 3)}{3(x - 1)(x + 1)}$
To divide one fraction by another, multiply the first fraction by the reciprocal of the second fraction:
$\dfrac{5(x + 3)}{10(x - 1)} \cdot \dfrac{3(x - 1)(x + 1)}{(x + 3)(x + 3)}$
Multiply numerator with numerator and denominator with denominator:
$\dfrac{5(x + 3) \cdot 3(x - 1)(x + 1)}{10(x - 1) \cdot (x + 3)(x + 3)}$
Cancel common terms in the numerator and denominator:
$\require{cancel}
\dfrac{\cancel{5}\cancel{(x + 3)} \cdot 3\cancel{(x - 1)}(x + 1)}{^2\cancel{10(}\cancel{x - 1)} \cdot \cancel{(x + 3})(x + 3)}$
$=\dfrac{3(x + 1)}{2(x + 3)}$
Restrictions on $x$ occur when the value of $x$ makes the fraction undefined, which means that the denominator becomes $0$.
Set denominators equal to $0$ and solve for $x$ to find where the restrictions are:
First denominator:
$10x - 10 = 0$
$10x = 10$
$x = 1$
Second denominator:
$3x^2 - 3 = 0$
$3x^2 = 3$
$x^2 = 1$
Take the square root of each side of the equation:
$x = \pm 1$
Third denominator (from the reciprocal):
$x + 3 = 0$
$x = -3$
Restrictions: $x \ne -3, -1, 1$
