19 and 42 are coprime, so we can use the CRT right away. Start with
![x=19+42](https://img.qammunity.org/2020/formulas/mathematics/college/5s3wrzks7fulw61t2w2e5hx28tlhy47370.png)
Taken mod 42, we're left with a remainder of 19. Notice that
![19\cdot3\equiv57\equiv15\pmod{42}](https://img.qammunity.org/2020/formulas/mathematics/college/a0l1qe82y43y2j1dfv27znkn1kyyb5vpjc.png)
so we need to multiply the first term by 3 to get the remainder we want.
![x=19\cdot3+42](https://img.qammunity.org/2020/formulas/mathematics/college/71x3xd1l7l9bq0ptnnmk7yz03dy8aehis5.png)
Next, taken mod 19, we're left with a remainder of 4. Notice that
![42\cdot6\equiv252\equiv5\pmod{19}](https://img.qammunity.org/2020/formulas/mathematics/college/7waq4jph2ytd02mwhs0d39vftr37qkws98.png)
so we need to multiply the second term by 6.
Then by the CRT, we have
![x\equiv19\cdot3+42\cdot6\equiv309\pmod{42\cdot19}\implies x\equiv309\pmod{798}](https://img.qammunity.org/2020/formulas/mathematics/college/6kpbyuy3c4dwuctgyn93t1qwbgrtbtdhhc.png)
so that any solution of the form
is a solution to this system.