In a senior secondary school, 80 students play hockey or football. The number that play football is 5 more than twice the number that play hockey. If the 15 students play both games and every student in the school playsat least one game, find:

The number of students that play football;
The number of students that play football but not hockey;
The number of students that play hockey but not footbal.

U={80}

5 more than twice of the number of hockey players plays football.
Let the hockey players(H) be X
Football players(F) be 2X+5
To find X
2X+5=FnH'+15. Equation1 X=HnF'+15. Equation2
80=(2X+5-15)+15+(X-15)
X=30

From here substitute X into equation1 and 2 to get F only nd H only

And remember that F will become 2(30)+5=65

solve it for me

To get x = 30 just add the number of two games and the no that play both and equate it to no of the student, and no of the student is 80

X-15+15+(2x+5)-15=80
X+2x+5-15=80
3x-10=80
3x=80+10
3x=90
X=30

To find no. (ii) and (iii)
ii. n(F)only= total no of F-- none
And total no of F is 2x+5
=2(30)+5
=65
Therefore, n(F)only=65-15 =50

iii. n(H)only = total no of H -- none
And total no of H is x which is 30
Therefore, n(H)only=30-15
=15

number who play hockey ---- x

number who play football -- 2x+5

x + 2x+5 = 80
x = 25

so 25 play hockey, 55 play football, and 15 play both

now we can use a Venn diagram, filling in as follows:
15 in the intersection
55-15 or 40 in the football only part of the circle
25-15 or 10 in the hockey only part of the circle

All your questions can now be easily answered.

pls i really need the working and answer,can u help me with it

In a senior secondary school, 80 students play hockey or football. The number that play football is 5 more than twice the number that play hockey. If the 15 students play both games and every student in the school playsat least one game, find:

The number of students that play football;
The number of students that play football but not hockey;
The number of students that play hockey but not football. Full working sir

Total number of student =80

N(F n H)=15
N(F)=5+2x
N(H)=x
(I). The number that play football
To get that , we need to sum up all the n( F) and n(H) to 80
I.e 5+2x-15 +15+X-15=80
3x-10=80
3x=90
X=30
Therefore; n(F)=5+2x
=5+60=65.
(ii). The number that play football but not hockey
N(F n H')
To get that we need to subtract the intercept from the to ta number that play football .and the intercept is 15
Therefore; 65-15=50
(ii). The number that play hockey but not football
N(F' n H)
We already know that x =30 and the intercept is 15
Therefore ; 30-15=15.

H=x

F=5+2x
Therefore, x+5+2x+15=80
3x+20=80
3x=60
X=20

Proof: x+5+2x+15
20+5+40+15=80

I got lost on Alexander explanation starting from how to get x

this maths is confusing especially on how to get the number of those who played football when hockey is 5 and hockey only is 20. i.e H=5+2x+15+x=80. 5+15+2x+x=80 20+3x=80 3x=80-20 3x=60 hockey only=20 hockey=20-15 =5