VCE Methods Question Thread!

[scribble]

that shouldn't be incorrect at all.
you can do a quick check. If ian is 33, then liz being six years younger, would be 27. adding those ages together gives 60. your answer is fine.

[Only Cheating Yourself]

that shouldn't be incorrect at all.
you can do a quick check. If ian is 33, then liz being six years younger, would be 27. adding those ages together gives 60. your answer is fine.

Shit i forget to check… The book says Ian is 36 while liz is 30.

Thanks for the help!

[hobbitle]

Well clearly the book is wrong because 30 + 36 doesn't equal 60.

[brightsky]

I need some help
1.What is the best way to solve it? Q: linear simultaneous equations: mx-5y=10 and 3x-(m-2)y=6.
   a)Find the values of m, where m belongs to R, for which are infinitely many or no solutions.
   b)Find the unique solution for the equation in terms of m.

2. Do I need to learn how to solve the questions algebraically? My teacher said we don’t need to know it.
X+y+z=9
-x+2y-3z+-15
X+5y+3z=29
Thanks in advance.

1. You can either solve the system using matrices as suggested by Stick or work it out using a bit of logic. Each equation represents a line on a Cartesian plane. No solutions means the two lines do not intersect. How can that happen? Same gradient but different y-intercept. Unique solutions means the two lines intersect once. How can that happen? Different gradient. Infinitely many solutions means the two lines are identical. How can that happen? Same gradient and same y-intercept. Use these three pieces of information to solve the problem intuitively.

2. For systems of more than two linear equations, use the substitution method. Rearrange the first equation to make.z the subject then sub that into the other two equations. Now z is gone. Rearrange one of the new rquations to make y the subject then sub inti the other equation. Now y is gone. You only have x. Solve for x and then sub back to obtain y and z. There are other techniques you can use as well. For example, you can put the system in matrix form AX=B and then solve by finding the inverse of the 3 x 3 matrix A, i.e. X=A^(-1) B. You can also express the system as an augmented matrix and then reduce it to reduced row echelon form. Neither of these techniques are in Methods though, but they are good to know.

[smile+energy]

1. Do you know how to solve simultaneous equations using matrices? If you can but can't remember, here's a hint: it involves working out the determinant.

2. Not really, most of the time you get that you will have your calculator to help you. However, it isn't a bad idea to try it by hand.

Thanks for your hint, I get it now.

[smile+energy]

1. You can either solve the system using matrices as suggested by Stick or work it out using a bit of logic. Each equation represents a line on a Cartesian plane. No solutions means the two lines do not intersect. How can that happen? Same gradient but different y-intercept. Unique solutions means the two lines intersect once. How can that happen? Different gradient. Infinitely many solutions means the two lines are identical. How can that happen? Same gradient and same y-intercept. Use these three pieces of information to solve the problem intuitively.

2. For systems of more than two linear equations, use the substitution method. Rearrange the first equation to make.z the subject then sub that into the other two equations. Now z is gone. Rearrange one of the new rquations to make y the subject then sub inti the other equation. Now y is gone. You only have x. Solve for x and then sub back to obtain y and z. There are other techniques you can use as well. For example, you can put the system in matrix form AX=B and then solve by finding the inverse of the 3 x 3 matrix A, i.e. X=A^(-1) B. You can also express the system as an augmented matrix and then reduce it to reduced row echelon form. Neither of these techniques are in Methods though, but they are good to know.

Thanks for your help . I would try different methods to solve these questions.

[Only Cheating Yourself]

At the movies a discount ticket sells for $5 off the normal price.  If four discount tickets and three full price thickets cost $92, how much does each type of ticket cost?

Let x=discounted price.

so x-5+4x+3=92
5x=94

And thats obviously in correct, i need help again please.

Thanks

[Sup]

At the movies a discount ticket sells for $5 off the normal price.  If four discount tickets and three full price tickets cost $92, how much does each type of ticket cost?

Let x=discounted price.

so x-5+4x+3=92
5x=94

And thats obviously in correct, i need help again please.

Thanks

So let x= normal price of ticket.

Therefore the discount price would be x-5.

"If four discount tickets and three full price thickets cost $92, how much does each type of ticket cost?"

So four discount tickets is 4(x-5). Three full price tickets = 3x.

Put all of it in an equation: 4(x-5) + 3x = 92
solve for x:

4x-20 + 3x = 92
7x = 112
x=16

Therefore the full price ticket costs $16.
The discount ticket is x-5 which is 16-5 = $11.

[Only Cheating Yourself]

So let x= normal price of ticket.

Therefore the discount price would be x-5.

"If four discount tickets and three full price thickets cost $92, how much does each type of ticket cost?"

So four discount tickets is 4(x-5). Three full price tickets = 3x.

Put all of it in an equation: 4(x-5) + 3x = 92
solve for x:

4x-20 + 3x = 92
7x = 112
x=16

Therefore the full price ticket costs $16.
The discount ticket is x-5 which is 16-5 = $11.

Thanks i knew we times 4 with something but why we do we times 4(x-5)?

[Sup]

Because the discount price is x-5, and we want four of them so it is 4 times (x-5) which is 4(x-5).

[Only Cheating Yourself]

At the agricultural show, a certain ride offers a discount of $1 on the full price of the ride when 10 or more tickets are purchased.  If 20 discount tickets cost the same as fifteen full price tickets, find the cost of each type of ticket.

let x=full price
discounted price=x-1


Far out!!!!!!!! Im fricken lost!

[e^1]

At the agricultural show, a certain ride offers a discount of $1 on the full price of the ride when 10 or more tickets are purchased.  If 20 discount tickets cost the same as fifteen full price tickets, find the cost of each type of ticket.

let x=full price
discounted price=x-1


Far out!!!!!!!! Im fricken lost!

"If 20 discount tickets cost the same as fifteen full price tickets"

[Orb]

"If 20 discount tickets cost the same as fifteen full price tickets"

^^yup, then just solve.

so 20x - 20 = 15x
5x = 20
x = 4

Each ticket costs $4

[Only Cheating Yourself]

What does CAS stand for?  Calculus something?

[alchemy]

What does CAS stand for?  Calculus something?

Computer Algebra System