VCE Methods Question Thread!

[Nato]

i seem to be getting this one wrong:
can anyone state the transformations to make the first equation into the second:

[brightsky]

1. Dilate by a factor of 1/2 from the x-axis. y = 1/(3-x)+2.
2. Reflect in the y-axis. y = 1/(x+3) + 2.
3. Translate 3 units in the positive direction of the x-axis. y = 1/x + 2.
4. Translate 2 units in the negative direction of the y-axis. y = 1/x.

This is only one possible sequence. There are other possibilities as well.

[Homer]

dilation of 1/2 from the x-axis
reflection in the y-axis
translation of 2 units down and 3 units right

[Nato]

1. Dilate by a factor of 1/2 from the x-axis. y = 1/(3-x)+2.
2. Reflect in the y-axis. y = 1/(x+3) + 2.
3. Translate 3 units in the positive direction of the x-axis. y = 1/x + 2.
4. Translate 2 units in the negative direction of the y-axis. y = 1/x.

This is only one possible sequence. There are other possibilities as well.

how about:
dilated by factor of 2 from x-axis
reflection in y-axis
translation 3 units in positive direction of x-axis
translation 4 units in positive direction of y-axis

[Homer]

how about:
dilated by factor of 2 from x-axis
reflection in y-axis
translation 3 units in positive direction of x-axis
translation 4 units in positive direction of y-axis

thats for going from 1/x to 2/3-x +4

[Bluegirl]

Stuck, again.
Have to solve the pairs of simultaneous equations.

Do I transpose it first or make a common factor?

3/2x - y= 4

1/2x + 3/4y = 10

And this one:

x/3 + y/2 = 11

x- y/3 = 22

Thankyou

I'm still doing something wrong

[clueless123]

I'm still doing something wrong

Substitution approach

      -----[1]
     -----[2]

from [1];





Substitute into [2]
 
 

expand, solve for x, then plug x back into [1]

[Nato]

I'm still doing something wrong

well for the first one:
you can transpose the equation into:
which you can substitute into
but i would recommend elimination, as it may make the process a little easier.

you can do this by multiplying the whole by and by doing so you'll get
and the other equation was   these are our simultaneous equations

now to find x, you have to eliminate y, which you can do by adding/subtracting on the the equations from the other. note: you do this to  'eliminate' the y terms, so you're left with x. after you get your x-value, sub this back into any of the equations and find y.

[Bluegirl]

Thanks everyone, finally got the answers.
How do I go about finding a value of k for which simultaneous equations have no solution or an infinite number of solutions?

[Nato]

Thanks everyone, finally got the answers.


How do I go about finding a value of k for which simultaneous equations have no solution or an infinite number of solutions?

is this with matrices?

[Bluegirl]

is this with matrices?

I worked it out, thanks.

Although matrices is coming up in the next few questions and I've never done it before.
Hopefully I understand it

[Bluegirl]

I feel like I'm asking too many questions.

A parabola passes through the points (-1,11) and (2,5) has equation y=x^2 + bx + c. Use simultaneous equations to find the values of b and c.

Not sure how to start

[Eugenet17]

I feel like I'm asking too many questions.

A parabola passes through the points (-1,11) and (2,5) has equation y=x^2 + bx + c. Use simultaneous equations to find the values of b and c.

Not sure how to start

Start by 11=(-1)^2+b(-1)+c [1] and 5=(2)^2+b(2)+c [2]
From here it's just regular simulataneous equations

[Bluegirl]

Start by 11=(-1)^2+b(-1)+c [1] and 5=(2)^2+b(2)+c [2]
From here it's just regular simulataneous equations

Was doing that but subbed in the wrong numbers haha. Woops
Thankyou!

[Only Cheating Yourself]

Find the exact so solution(if they exist) or prove there are no solutions

5(4x+3)=(4x+3)^2+9

I can use the substation method i.e (4x+3) but theres an = sign so i now don't know how to use it…