Principle of inclusion exclusion

Question

In a class of 30 children, 20 studied Portuguese, 14 studied English and 10 studied French. If 8 study none of these 3 languages ​​and none study the 3 languages, how many children study English and French?

How do I approach this problem?

Answer: 2

Answer 1

22 students studied at least one language

Let the number who studied English but neither French nor Portuguese be denoted by $E$, the number who studied only French be $F$, and the number who studied Portuguese be $P$.

Let the number who studied both English and French (but necessarily not Portuguese) be denoted by $EF$, and so forth. We then know that

$$ P+E+F+PE+PF+EF = 22 \\ P+PE+PF = 20\\ E+PE+EF = 14 \\ F+EF+PF = 10 $$Add the last three equations and subtract from that the first equation, to get $$ PE + PF + EF = (20+14+10)-22 = 22\\ \implies PF = 22-PE-EF $$ Then use that in the fourth equation to get $$ F + EF + 22 - PE -EF = 10 \\ \implies PE = F + 12 $$ Plug that into the third equation to get $$ E + (F+12) + EF = 14 \implies E+F+EF = 2 $$ The problem is flawed, in that $2$ students took English or French or both, so the number that took both English and French has to be negative.