Algebra and Trigonometry 10th Edition

$x=\frac{11}{6}$
$x\ne\pm0.5$ because the denominators cannot be equal to $0$. After multiplying both sides by $(2x+1)(2x-1):$ $7(2x-1)-8x(2x+1)=-4(2x-1)(2x+1)\\14x-7-16x^2-8x=-4(4x^2-1)\\-16x^2+6x-7=-16x^2+4\\6x=11\\x=\frac{11}{6}$ Check by plugging in the solution: $\frac{7}{2\frac{11}{6}+1}-\frac{8\frac{11}{6}}{2\frac{11}{6}-1}=\frac{7}{\frac{11}{3}+1}-\frac{\frac{44}{3}}{\frac{11}{3}-1}=\frac{7}{\frac{14}{3}}-\frac{\frac{44}{3}}{\frac{8}{3}}=\frac{21}{14}-\frac{44}{8}=\frac{21}{14}-\frac{44}{8}=1.5-5.5=-4$ Thus, it is a good solution.