The old machine's rate is $\displaystyle\frac{{24}}{{3}}\usepackage{cmbright}\Huge\$ or 8 widgets per hour. The new machine's rate is $\displaystyle\frac{{48}}{{4}}\usepackage{cmbright}\Huge\$ or 12 widgets per hour. Their combined rate is 20 widgets per hour. So, if $\displaystyle{t}=\usepackage{cmbright}\Huge\$ the amount of time, in hours, needed for the two machines to produce 120 widgets, then:

$\displaystyle{20}=\frac{{120}}{{t}}\usepackage{cmbright}\Huge\$ , so $\displaystyle{20}{t}={120}\usepackage{cmbright}\Huge\$ , and $\displaystyle{t}={6}\usepackage{cmbright}\Huge\$