Solve For Y; 2x-5y=7?

Add 5y-7 to both sides, then divide by 5.
2x-7 = 5y
y = (2/5)x - (7/5)

I thought there was a way to get a real number answer as long as you had enough information.

I have played a game a total of 953 times. Winning 876, I have lost 77 games giving me a win rate of ~92%.

I want to know how many more games I have to win in a row to achieve 99%.

I think this is actually monomial because I have two unknown factors but they will be the same: How many more games to be played.

Based on the old cross-multiply method of getting percentages.

In this case the cross multiply would

99/100 = (current games lost plus unknown additional games played) divided by (current games played plus x unknown additional games played)

The number of additional games played will be the same, hence x.

So:

99/100 = (77+x) / (876 + x)

So Cross multiply
99 (876+x) = 100(77+x)
86724 + 99x = 7700 + 100x

What's next?
Subtract 7700 from both sides to simplify?

79024 + 99x = 100x ?

Can I subtract 99x from both sides?

79024 = x?

This works when I run the results through to see if I get 99%. But it feels wrong. It's been 40 years since I did algebra. Anyone want to set an old man straight?