Question

ALL SMART PEOPLE IN MATH PLEASE HELP ME I AM VERY SICK

Answer 1

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Let to be x represent the entry fee

Also,

Let to be y represent the cost of each ticket in dollars..

ACCORDING TO THE QUESTION

Gwen pays the entry fee and buys 10 tickets. It costs her a total of $30

then,

Equation be formed:-
x+10y=30

Again ,
According to the question

Tristan pays the entry fee and buys 15 tickets. It costs him a total of $40.

Equation be formed:-

x+15y=40

Now

Equation be formed

Solve the Equation by Elimination method

x+10y=30

x+15y=40

_________
-5y=-10

y=-10/-5

y=2

Now

Putting the value of y in another equation

x+15y=40

x+15(2)=40

x+30=40

x=40-30

x=10

Then,

Entry fee=10

Cost of each ticket in Dollars=2


x+10y=30

Now putting the value of x and y in thirds Equation..


then,

We get


Note:-. It's additional Number of Value


x+10y=30



10+10(2)=30

10+20=30

30=30


Verified...

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Answer 2

Hope u like my process
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=> Let the cost of the entry fee be $ x

=> Let the cost of the tickets be $ y
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By problem,
=-=-=-=-=-=-=-=-
$\: \: \: \: \: \: \: \: \: \bf\underline{ \: \: part \: \: a \: \: }$

=> Gwen pays entry fee and buys 10 tickets and it costs her $ 30

So, x + 10y = 30 ____(1)
________________________
$\: \: \: \: \: \: \: \: \: \bf \underline{ \: \: \: part \: \: \: b \: \: \: }$

=> Tristan pays the entry fee and buys 15 tickets and it costs her $ 40

So, x + 15y = 40 _____(2)

________________________
Thus,

Comparing equation (1)&(2)
=-=-=-=-=-=-=-=--=-=-=-=-=-=-=
$= > x + 15y = 40 \\ \underline{- \: \: \: \: \: x + 10y = 30 \: \: \: } \\ = > \bf \: 5y = 10 \\ \\ = > \bf \: y = \frac{10}{5} = 2$

Thus, cost of a ticket is $ 2.

Cost of entry fee = $ 10
__________________________
=> Keith paid the same as Gwen so,

=> x + 10 y = 30...

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Hope this is ur required answer

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