simultaneous equation :3ab+b=10 and 2a+4b-0

Rearrange first equation: b(3a+1)=10, so b=10/(3a+1);

the second equation has no equals sign, and if the minus sign is supposed to be an equals sign, b=-a/2, so -a/2=10/(3a+1).

-3a^2/2-a/2=10 or 3a^2+a+20=0 has no real solution.

The question needs to be checked, corrected and resubmitted. There are complex roots, but no real roots.