algebra precalculus - Two workers can do a job, first one for 8 days and the second for 12

If they work seperately, the first one does the job for 8 days, and the other for 12 days.

After first worker worked for 2 days, he got the help of the second worker, for how many days they finished the job.

My try
$$\frac {8}{x}=1$$
And
$$\frac {12}{y} =1$$

But now I'm stuck...

Answer

Notice

job of first worker in one day $$=\frac{1}{8}$$
job of second worker in one day $$=\frac{1}{12}$$

Now, let the job be finished in $x$ days when both the workers work together.

The job left out (remaining) after first worker works for 2 days $$=1-2\times \frac{1}{8}=\frac{3}{4}$$

The left out (remaining) $\frac{3}{4}$ of the complete job is to be finished by both the workers in $x$ days then we have $$x\left(\frac{1}{12}+\frac{1}{8}\right)=\frac{3}{4}$$
$$x\left(\frac{5}{24}\right)=\frac{3}{4}$$

$$x=\frac{3}{4}\times \frac{24}{5}$$
$$x=\frac{18}{5}=3.6$$
Thus the both the workers together will finish the left out (remaining) job in $3.6\approx 4\ days$