## College Algebra (6th Edition)

a. Restriction values $: 0$ b. Solution set $:\{ 3 \}$
$\frac{5}{2x} - \frac{8}{9} = \frac{1}{18} - \frac{1}{3x}$ a. $x =0$ makes the denominator zero. So, $x \ne 0.$ b. $\frac{5}{2x} - \frac{8}{9} = \frac{1}{18} - \frac{1}{3x};x \ne 0;$ Take LCD on both sides. $\frac{45-16x}{18x} = \frac{x-6}{18x} ;x \ne 0;$ Multiply both sides by $18x$ to clear fraction part. $18x(\frac{45-16x}{18x}) = 18x(\frac{x-6}{18x}) ;x \ne 0;$ $45-16x = x-6$ $45+6 = x+16x$ $51 = 17x$ $x = \frac{51}{17}$ $x = 3$ is not part of the restriction values. So, Solution set $:\{ 3 \}$