# 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)
thanked the writer.
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?
thanked the writer.
Basically follow the same instructions for your previous question, then subtracting "7" unilaterally to isolate the "Y" variable, then divide both side by "5"
thanked the writer. 