## Intermediate Algebra (6th Edition)

$x=\dfrac{4}{7}$
Raising both sides of the given equation, $\sqrt[]{7x-4}=\sqrt[]{4-7x} ,$ to the second power, then the solution/s is/are \begin{array}{l}\require{cancel} 7x-4=4-7x \\\\ 7x+7x=4+4 \\\\ 14x=8 \\\\ x=\dfrac{8}{14} \\\\ x=\dfrac{\cancel{2}\cdot4}{\cancel{2}\cdot7} \\\\ x=\dfrac{4}{7} .\end{array} Upon checking, $x=\dfrac{4}{7}$ satisfies the original equation.