## College Algebra (10th Edition)

{$-\frac{14}{139}$}
$\frac{8w+5}{10w-7}=\frac{4w-3}{5w+7}$ Where $w\ne\frac{7}{10},w\ne-\frac{7}{5}$ $(8w+5)(5w+7)=(4w-3)(10w-7)$ Use cross multiplication. $40w^2+56w+25w+35=40w^2-28w-30w+21$ Multiply. $40w^2+81w+35=40w^2-58w+21$ Combine like terms. $40w^2-40w^2+81w+35=40w^2-40w^2-58w+21$ Subtract $40w^2$ from both sides. $81w+35=-58w+21$ $81w+58w+35=-58w+58w+21$ Add 58w to both sides. $139w+35=21$ $139w+35-35=21-35$ Subtract 35 from both sides. $139w=-14$ $\frac{139w}{139}=\frac{-14}{139}$ Divide both sides by 139 $w=-\frac{14}{139}$ The solution is {$-\frac{14}{139}$}