## Precalculus: Mathematics for Calculus, 7th Edition

$x=-\dfrac{1}{2}$
$\sqrt{3}x+\sqrt{12}=\dfrac{x+5}{\sqrt{3}}$ We can take the denominator $\sqrt{3}$ to multiply to the left side of the equation: $\sqrt{3}\sqrt{3}x+\sqrt{3}\sqrt{12}=x+5$ Let's remember that $\sqrt{a}\sqrt{b}=\sqrt{ab}$. Apply this property to evaluate the products present in the equation: $\sqrt{(3)(3)}x+\sqrt{(3)(12)}=x+5$ $\sqrt{9}x+\sqrt{36}=x+5$ Simplify and solve for $x$: $3x+6=x+5$ $3x-x=5-6$ $2x=-1$ $x=-\dfrac{1}{2}$