we need to find y games won OUT OF x games played last season, with the condition of 0 tied games; lost games will be (x-y)

st(1) team lost two more games last season, won 20%(and lost 80%) of all games -> (x-y)+2=0.8 and (y-2)/x=0.2 By solving the system of two equations we can find y. Not Sufficient

{(x-y+2)/x=0.8 -----------------> xy=0.2x+2
{(y-2)/x=0.2 ---> y=0.2x+2 <>

st(2) team won three more games, lost 30% (and won 70%) of all games -> (y+3)/x=0.7 and (x-y)-3=0.3. Again by solving the system of two equations we can find y. Not Sufficient

{(y+3)/x=0.7 --> y+3=0.7x
{(x-y-3)/x=0.3 -----------> y+3=0.7x;

combined st(1&2): y+3=0.7x U y=0.2x+2 Sufficient

MI3 wrote:A certain football team played x games last season, of which the team won exactly y games. If tied games
were not possible, how many games did the team win last season?
(1) If the team had lost two more of its games last season, it would have won 20 percent of its games for
the season.
(2) If the team had won three more of its games last season, it would have lost 30 percent of its games
for the season.

Is the answer C ?

Statement 1: Losing 2 more games = winning 20%
y-2 = .2x
The equation above has an infinite number of solutions.
Insufficient.

Statement 2: Winning 3 more games = losing 30% = winning 70%
y+3 = .7x
The equation above has an infinite number of solutions.
Insufficient.

Statements 1 and 2 together:
2 variables, 2 distinct linear equations.
Sufficient.

The correct answer is C.

If this were a PS question that we needed to solve:
Dividing the first equation by the second, we get the following ratio:
(y-2)/(y+3) = 2/7
7y-14 = 2y+6
5y = 20
y = 4.

Thanks for explenation

Thank you Mitch.