VCE Methods Question Thread!

[Sine]

There's this question that I am struggling to complete which is in the quadratics section of the textbook. Can someone please solve this using some form of quadratics and explain each step that you take to solve it. Thank you.

Tickets for a concert are available at two prices. The more expensive ticket is $30 more than the cheaper one. Find the cost of each type of ticket if a group can buy 10 more of the cheaper tickets than the expensive ones for $1800.

This quesiton has been asked and answered before
https://atarnotes.com/forum/index.php?topic=177148.0

[PolySquared]

This quesiton has been asked and answered before
https://atarnotes.com/forum/index.php?topic=177148.0

The answer that was provided in that thread utilised simultaneous equations only and did not involve quadratics. I am looking for an answer where the solution process involved some form of quadratics.

[VanillaRice]

There's this question that I am struggling to complete which is in the quadratics section of the textbook. Can someone please solve this using some form of quadratics and explain each step that you take to solve it. Thank you.

Tickets for a concert are available at two prices. The more expensive ticket is $30 more than the cheaper one. Find the cost of each type of ticket if a group can buy 10 more of the cheaper tickets than the expensive ones for $1800.

Let x be the cost of the cheaper ticket, and n be the number of expensive tickets purchased to give a cost of $1800.
x+30 will therefore be the cost of the more expensive ticket, and n+10 will be the number of cheap tickets.
The number and cost of each ticket can be related by the equations:

Rearrange equation (2) to give

Substitute this into equation (1), and you will be able to rearrange the equation to give you a quadratic. I'll let you attempt the rest.

Hope this helps Post with your progress if you get stuck.

[PolySquared]

Let x be the cost of the cheaper ticket, and n be the number of expensive tickets purchased to give a cost of $1800.
x+30 will therefore be the cost of the more expensive ticket, and n+10 will be the number of cheap tickets.
The number and cost of each ticket can be related by the equations:

Rearrange equation (2) to give

Substitute this into equation (1), and you will be able to rearrange the equation to give you a quadratic. I'll let you attempt the rest.

Hope this helps Post with your progress if you get stuck.

Thanks for your reply. I don't understand how you got the second equation, could you please explain why the number of expensive tickets bought multiplied by the cost is equal to 1800?

[VanillaRice]

Thanks for your reply. I don't understand how you got the second equation, could you please explain why the number of expensive tickets bought multiplied by the cost is equal to 1800?

Your question stated "Find the cost of each type of ticket if a group can buy 10 more of the cheaper tickets than the expensive ones for $1800"
In the cases of both the cheap and expensive tickets, we are spending $1800. If n is the number of expensive tickets bought to give a total cost of $1800, then multiplying by the cost of the expensive ticket, n(x+30) will give a total of $1800. It's the same as saying that if I buy five $2 pens, I have to spend $10 in total.

Hope that makes sense

[PolySquared]

Your question stated "Find the cost of each type of ticket if a group can buy 10 more of the cheaper tickets than the expensive ones for $1800"
In the cases of both the cheap and expensive tickets, we are spending $1800. If n is the number of expensive tickets bought to give a total cost of $1800, then multiplying by the cost of the expensive ticket, n(x+30) will give a total of $1800. It's the same as saying that if I buy five $2 pens, I have to spend $10 in total.

Hope that makes sense

Thank you so much, it makes sense.

[TheAspiringDoc]

Hi,

Solve the simultaneous equations 2x - 3y = 4 and x + ky = 2, where k is a constant

So putting it in to the simultaneous equations solver on my CASIO fx-CP400, I get:
x = 2, y = 0
Which is correct
However the correct answer should also include that k =/= -3/2, which my CAS didn't give.

Does this mean I should have solved it by hand? Or is there a graphical method on the calculator that will give the full answer?

[snowisawesome]

Is it ok if I sometimes do the exercices in the textbook in reverse order?
Like I do exercise 3.5 first, because it's easier, then do exercise 3.3
Is this ok?

Also, is y = 2+((1/x-3)) a one-to-one function and if so, how can you tell that it is?

Thanks

[keltingmeith]

Is it ok if I sometimes do the exercices in the textbook in reverse order?
Like I do exercise 3.5 first, because it's easier, then do exercise 3.3
Is this ok?

Also, is y = 2+((1/x-3)) a one-to-one function and if so, how can you tell that it is?

Thanks

There's no reason why that would be a bad idea.

You can tell by doing a horizontal line test. If its inverse is a function, then what you have is a one-to-one function.

[snowisawesome]

There's no reason why that would be a bad idea.

You can tell by doing a horizontal line test. If its inverse is a function, then what you have is a one-to-one function.

what do you mean by this?

[Bell9565]

what do you mean by this?

A horizontal line test is that if you draw a horizonal line (eg y=6 ect) in the plane, if it is a one to one function (or a many to one function) then it will only intersect once or not at all. If it were one to one it would also pass the vertical line test. 

[snowisawesome]

A horizontal line test is that if you draw a horizonal line (eg y=6 ect) in the plane, if it is a one to one function (or a many to one function) then it will only intersect once or not at all. If it were one to one it would also pass the vertical line test. 

So would I do the horizontal line test on the original graph or its inverse?

[keltingmeith]

So would I do the horizontal line test on the original graph or its inverse?

The original.

Your textbook should explain this in more detail.

[Bell9565]

So would I do the horizontal line test on the original graph or its inverse?

Also I might add, if it is a function then the inverse will not exist if it does not pass both the horizontal/vertical line tests (thus being one to one). If you were to do the line test on an inverse function, it will always pass both, as its original function would be one to one. If it doesn't pass then it's not an inverse function (even though the 'rule' may exist it, it is still not an inverse function)

[VanillaRice]

Hi,
So putting it in to the simultaneous equations solver on my CASIO fx-CP400, I get:
x = 2, y = 0
Which is correct
However the correct answer should also include that k =/= -3/2, which my CAS didn't give.

Does this mean I should have solved it by hand? Or is there a graphical method on the calculator that will give the full answer?

k = -3/2 will give two equations which are exactly the same (except the coefficients differ by a factor of 2). If this was an exam, I would speculate that the domain for k would already be given. Otherwise, obtaining this answer just requires some thought about the relationship between two linear equations containing variables (in particular, how they can have infinite, unique or no solutions)