Help with math????

Question

1)Tickets for a concert sold for $11 and $8. If 600 tickets were sold for one evening and the the receipts are $5550, how many of each kind of tickets were sold?

2)Solve by substitution--
x+2y=10
y=2x
and
2x+y=9
x-3y=22

If some one could help so I understand

thanks

Answer 1

Hi,

1)Tickets for a concert sold for $11 and $8. If 600 tickets were sold for one evening and the the receipts are $5550, how many of each kind of tickets were sold?

Let x = number of $11 tickets
Let y = number of $8 tickets

x + y = 600 <-- total number of tickets
11x + 8y = 5550 <-- total cost of tickets

Multiply the top equation by -8. Then add equations together. "y" will drop out and then solve for x.

-8(x + y = 600)
11x + 8y = 5550

-8x - 8y = -4800
11x + 8y = 5550
------------------------
3x = 750
x = 250, so there are 250 $11 tickets

If x = 250, then 250 + y = 600
y = 350, so there are 350 $8 tickets


2)Solve by substitution--
x+2y=10
y=2x

Replace y in the first equation with 2x.

x + 2(2x) = 10
x + 4x = 10
5x = 10
x = 2

Find y by plugging in x = 2.

If y = 2x, then y = 2(2) so y = 4.

The answer is (2,4).


2)Solve by substitution--

2x+y=9
x-3y=22

Solve the 2nd equation for x.

x - 3y = 22
x = 3y + 22
Substitute this expression into the first equation in place of x.

2x + y = 9
2(3y + 22) + y = 9
6y + 44 + y = 9
7y + 44 = 9
7y = -35
y = -5

Substitute -5 for y and solve for x.

2x + y = 9
2x - 5 = 9
2x = 14
x = 7

So the answer is (7,-5).

I hope those help you!! :-)

Answer 2

1)Tickets for a concert sold for $11 and $8. If 600 tickets were sold for one evening and the the receipts are $5550, how many of each kind of tickets were sold?

11x + 8y = 5550
x + y = 600

x = 600 - y

11(600 - y) + 8y = 5550
6600 - 11y + 8y = 5550
-3y = -1050
y = 350

x + y = 600
x = 600 - y
x = 600 - (350)
x = 250

350 tickets were $8
250 tickets were $11



2)
x+2y=10
y=2x

x + 2(2x) = 10
x + 4x = 10
5x = 10
x = 2

y = 2x
y = 2(2)
y = 4


and

2x+y=9
x-3y=22

x - 3y = 22
x = 22 + y

2x + y = 9
2(22 + y) + y =9
44 + 2y + y = 9
3y = -35
y = -35/3

x - 3y = 22
x = 22 + 3y
x = 22 + 3(-35/3)
x = 22 - 35
x = -13

Answer 3

For the tickets:

First let X = the number of $11 tickets

Then 600-X would be the number of $8 tickets

We know the total value was 5550$; so,

11X + 8 x [600-X] = 5550

11X + 4800 - 8X = 5550

11X -8X = 5550 - 4800

3X = 750

X = 250 so 250 11$ tickets and [600-250] 8$ tickets.

Check your answer: [250 x 11] + [350 x 8] = 5550

X + 2Y = 10; Y = 2X (substitute 2X for Y in 1st equation)

X +2(2X) = 10 >> X + 4X = 10 >>5X = 10 >>X =2

Answer 4

1)
11x + 8y = 5550
x + y = 600
y = 600 - x
substitute:
11x + 8(600 - x) = 5550
11x + 4800 - 8x = 5550
x = (5550 - 4800)/3
x = 250
y = 600 - 250 = 350

2)
x + 2y = 10
y = 2x
x + 2(2x) = 10
x = 10/5
x = 2
y = 2(2) = 4

3)
2x + y = 9
x - 3y = 22
x = 22 + 3y
2(22 + 3y) + y = 9
44 + 6y + y = 9
44 + 7y = 9
y = -5
2x - 5 = 9
x = 7

Answer 5

1) let number of 11 bucks tix to x and let number of 8 bucks tix to be y.

Total tickets sold is 600. So x+y=600. This is ur first equation.
Then total receipts is 5500. So 11x +8y=5500. This is ur second equation.

x +y=600
Therefore, x=600-y
Subtitute ur x into the second equation which you derived earlier one, and you'll have the answer.

2) Since y=2x, so
x+2y=10
x+2(2x)=10
x+4x=10
5x=10
x=2

x-3y=22
so, x=22+3y
Sub x=22+3y into 2x+y=9
You'll have 2(22+3y)+y=9
which is equal to 44+6y+y=9
7y=9-44
7y= -35
therefore, y=5

Answer 6

1) set it up like this-------------> 11x + 8y = $5550

You do it yourself.

2) 3x = 10

and

7y = -35

You should do these yourself...