Maths Question

1. Let y be the number of $20
y+y+5+0.5y+7=22
2y+0.5y+12 =22
2y+0.5y+12-12=22-12
2.5y=10
2.5y/2.5=10/2.5
y=4
.
. .David has$20x4
$10x9(4+5)
$50x9(4/2+7)

2. Let y be the number of student's tickets were sold
5y+7.5(450-y)=2600
5y+(450*7.5-450*y)=2260
5y+(3375-7.5y)=2260
5y+3375-7.5y=2260
5y-7.5y+3375-3375=2260-3375
-2.5y=-775
-2.5y/-2.5=-775/-2.5
y=310

.
. .310 student's tickets were sold
140(450-310)adult's tickets were sold

1) let x be $10, y be $20, z be $50

y+5=x---------(1)
1/2y+7=z------(2)
x+y+z=22------(3)

sub1,2 into3,

y+5+y+1/2y+7=22
2.5y=10
y=4

bcz
y+5=x
so x=9

bcz
x+y+z=22
so z=9

so$10=4 $20=9 $30=9

2)let a be the tickets of adult, k be the tickets of th kids

450-a=k---------------------------------(1)
(450-a)5+(450-k)7.5=2600------------(2)

sub 1 into 2
(450-a)5+(450-450+a)7.5=2600
2250-5a+7.5a=2600
2.5a=350
a=140

bsz
450-a=k
so k=310

so 310 tickets of each kid were sold

For the 1) question , we let the number of $20 note as "X" , then the number of $10 note is "X-5" and $50 note is "0.5X+7" , the equation is
X + X-5 + 0.5X + 7 = 22
=> 2.5X +2 = 22
=> 2.5X = 20
=> X = 8
So we have the answer the number of $20 note is 8 , the number of $10 is 3 (8-5) and the number of $50 note is 11 (0.5x8+7)

For the 2) , we let the number of of adult's ticket is X and the number of student ticket is Y , then we get
i) X + Y = 450
ii) 7.5X + 5Y = 2600
From equation i) , we get Y = 450 - X , we substitute it into ii) , then
7.5X + 5(450-X) = 2600
=> 7.5X + 2250 - 5X = 2600
=> 2.5X + 2250 = 2600
=> 2.5X = 350
=> X = 140
Then we get X = 140 and Y = 310 .