## Precalculus (6th Edition)

$\sqrt{x+7}+3=\sqrt{x-4}$ Square both sides of the equation: $(\sqrt{x+7}+3)^{2}=(\sqrt{x-4})^{2}$ $x+7+6\sqrt{x+7}+9=x-4$ Leave the square root term alone on the left side of the equation: $6\sqrt{x+7}=x-4-x-7-9$ $6\sqrt{x+7}=-20$ Take $6$ to divide the right side: $\sqrt{x+7}=-\dfrac{20}{6}$ $\sqrt{x+7}=-\dfrac{10}{3}$ Square both sides: $(\sqrt{x+7})^{2}=\Big(-\dfrac{10}{3}\Big)^{2}$ $x+7=\dfrac{100}{9}$ Solve for $x$: $x=\dfrac{100}{9}-7$ $x=\dfrac{37}{9}$ The value of $x$ found is not a solution for the equation, it has no solution.