Question 1184050
let x = number of girls.
let y = number of boys.


original ratio is x/y = 5/3.


second ratio is (x+3)/(y+5) = 7/5.


solve for x in the first ratio to get x = 5y/3.


solve for (x+5) in the second ratio to get (x+3) = 7*(y+5)/5.
simplify to get (x+3) = (7y+35)/5
multiply both sides of that equation by 5 to get 5x+15 = 7y+35
subtract 15 from both sides of that equation to get 5x = 7y+20
replace x with 5y/3 fom the first equation to get 5*5y/3 = 7y+20
simplify to get 25y/3 = 7y+20
multiply both sides by 3 to get 25y = 21y+60
subtract 21y from both sides to get 4y=60
solve for y to get y = 60/4 = 15


the original ratio is x/y = 5/3
when y = 15, the ratio becomes x/15 = 5/3
solve for x to get x = 5*15/3 = 5*5 = 25


you have x = 25 and y = 3
x is the number of girls.
y is the number of boys.


the original ratio is 25/15 = 5/3.
the revised ratio is 28/20 = 14/10 = 7/5.


both ratios are good, confirming the solution is correct.


the solution is there were 25 girls in the club originally.


i did a graphical solution and came up with the same answer.
this is what it looks like.


<img src = "http://theo.x10hosting.com/2021/090101.jpg" >