1. ## basbell problem

Q: if a basebell nine should have play 2 games more and win both, it will have won 2/3 of the games played. However, if it should have plAYED 7 win 4 of them, it will then have won 3/5 of the games played. How many games has it so far played and how many has it won?

my attempt:
Q: After going a certain distance in an automoble, a driver found that if he had gone 3 miles an hour faster, he would have traveled the distance in 1 hour less time. and that if he had gone 5 miles an hour faster, he would have gone the distance in 1 hour and 30 minutes less.
What was the distance?

my attempt:

$x+2 = \frac{2}{3}y$

$x+ 4 = \frac{3}{5}y -3$

$7 = \frac{3}{5}y -x (2)$
$2 = \frac{2}{3}y -x (1)$

(2) - (1)

produces $5= - \frac{1}{15}y$

this will cause a negative number hence why i believe i am wrong

please how do i fix this

2. ## Re: basbell problem

Originally Posted by bigmansouf
Q: if a basebell nine should have play 2 games more and win both, it will have won 2/3 of the games played. However, if it should have plAYED 7 win 4 of them, it will then have won 3/5 of the games played. How many games has it so far played and how many has it won?

my attempt:
Q: After going a certain distance in an automoble, a driver found that if he had gone 3 miles an hour faster, he would have traveled the distance in 1 hour less time. and that if he had gone 5 miles an hour faster, he would have gone the distance in 1 hour and 30 minutes less.
What was the distance?
???

3. ## Re: basbell problem

Originally Posted by bigmansouf
Q: if a basebell nine should have play 2 games more and win both, it will have won 2/3 of the games played. However, if it should have plAYED 7 win 4 of them, it will then have won 3/5 of the games played. How many games has it so far played and how many has it won?

my attempt:

Q: After going a certain distance in an automoble, a driver found that if he had gone 3 miles an hour faster, he would have traveled the distance in 1 hour less time. and that if he had gone 5 miles an hour faster, he would have gone the distance in 1 hour and 30 minutes less.
What was the distance?

my attempt:

$x+2 = \frac{2}{3}y$
$x+ 4 = \frac{3}{5}y -3$

$7 = \frac{3}{5}y -x (2)$
$2 = \frac{2}{3}y -x (1)$
So this is the first problem, right? Either place the solution immediately after the problem or label them! Also your solution makes no sense because you haven't said what "x" and "y" represent! Since the problem talks about "2 more games" I guess that "x" is the number of game won so far. But what is "y"? It would be reasonable to make "y" the number of game played so far but then, after playing two more games that should be "y+ 2" on the right, not "y". But then where did the "x+4= (3/5)y- 3 come from?

(2) - (1)

produces $5= - \frac{1}{15}y$

this will cause a negative number hence why i believe i am wrong

please how do i fix this[/QUOTE]
I cannot understand what you have done or where you got those equations. In the future, if you post problems here, tell us what you letters represent and how you got the equations you wrote! (If I were you instructor, I would expect you to do that in order to get full credit for a problem.)

Suppose there are m games played so far, n of them were won. With 2 more games, both won, there are m+ 2 games, n+ 2 of them won. So they will have won $\frac{n+2}{m+2}= \frac{2}{3}$. I presume that "played 7 win 4 of them" means "more than the original and n" so if they played 7 more game, rather than only 2, and won 3 of them they will have won $\frac{n+ 4}{m+7}= \frac{3}{5}$ of their games. That gives two equations to solve for m and n.

4. ## Re: basbell problem

Originally Posted by skeeter
???
sorry