## College Algebra (6th Edition)

a. Restricted Value : $0$ b. Solution Set : $\{ \frac{5}{12} \}$
$\frac{5}{x} = \frac{10}{3x} + 4$ a. It $x=0$, it makes the denominator zero, therefore, $x=0$ is the restricted value. b. $\frac{5}{x} = \frac{10}{3x} + 4 ; x \ne 0;$ Take LCD at the right hand side. $\frac{5}{x} = \frac{10+12x}{3x} ; x \ne 0;$ Multiply both sides by $3x$ to clear fractional part. $3x(\frac{5}{x}) = 3x(\frac{10+12x}{3x}) ; x \ne 0;$ $15 = 10+12x$ $15-10=12x$ $12x = 5$ $x= \frac{5}{12}$ $\frac{5}{12}$ is not the restricted value. Solution Set : $\{ \frac{5}{12} \}$