The swim and diving clubs at riverdale high school have a total of 55 members and no student is a member of both teams. of the swim team members are seniors and of the diving team members are seniors. if there are 13 seniors in the two clubs, how many members does each club have? let x represent the total number of swim club members and let y represent the total number of diving club members. the equation that represents the total number of members is . the equation that represents the total number of seniors is . the diving club has more members than the swim club.

The first question x+y=55
The second question (1/3)x+(1/5)y=13
The third question 25

We get two equations from the information given:
(1/3)x + (1/5)y = 13
and
x + y = 55
x being swim club members and y being diving club members.

To get additive inverses, we need to multiply both sides by two different numbers. We are going to solve for x.

We multiply the first equation by 10 and the second by -2:
(1/3)x + (1/5)y = 13 multiplied by 10
x + y = 55 multiplied by -2
Now we have:
10/3x + 2y = 130
-2x - 2y = -110
When we add these two equations together, the 2y from the first equation and the -2y from the second equation cancel out, leaving us with:
4/3x = 20
We multiply both sides by 3 to get rid of the fraction, giving us:
4x = 60
We divide both sides by 4 to isolate the variable, giving us:
x = 15
This is the number of swim club members. We are trying to find how many more members the diving club has compared to the swim club, so we must find y. Just substitute 15 for x in any equation, I will substitute it into the second equation and solve:
15 + y = 55
y = 55 - 15
y = 40
You might think we are finished, but we have one more step. They are asking for  how many more members the diving club has compared to the swim club. To find this, we subtract y and x.
40-15
The answer is 25 more diving club members than the swim club!

Idon’t exactly understand but i think dbc=rst