Intermediate Algebra (6th Edition)

$\frac{x-1}{2x-3}$ where $x\ne\frac{3}{2}$
We are given that $f(x)=x-1$ and $g(x)=2x-3$. We know that $(\frac{f}{g})(x)=\frac{f(x)}{g(x)}$. Therefore, $(\frac{f}{g})(x)=\frac{f(x)}{g(x)}=\frac{x-1}{2x-3}$. However, the denominator cannot be equal to 0. Therefore, $2x-3\ne0$. Add 3 to both sides. $2x\ne3$ Divide both sides by 2. $x\ne\frac{3}{2}$