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October 30, 2025, 11:06:46 am

Author Topic: VCE Methods Question Thread!  (Read 5766029 times)  Share 

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Conic

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Re: VCE Methods Question Thread!
« Reply #3330 on: December 23, 2013, 06:56:33 pm »
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so when your moving a number or expression over an = sign the signs change but in this case 2y just moves over but the signs say the same?
Yes the sign changes when you move it, but the 2y isn't moving in this case.



Move the -3x+6 to the left:



2012-13: VCE at Parade College (Chemistry, English, Mathematical Methods, Physics and Specialist Mathematics).
2014-16: Bachelor of Science at La Trobe University (Mathematics and Statistics).
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Only Cheating Yourself

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Re: VCE Methods Question Thread!
« Reply #3331 on: December 23, 2013, 07:00:17 pm »
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Yes the sign changes when you move it, but the 2y isn't moving in this case.



Move the -3x+6 to the left:





2y+3x-2=0 i need to now put that into ax+by+c=0.  So when i move the x and y around the signs say the same?
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Conic

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Re: VCE Methods Question Thread!
« Reply #3332 on: December 23, 2013, 07:07:23 pm »
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2y+3x-2=0 i need to now put that into ax+by+c=0.  So when i move the x and y around the signs say the same?
Yes. 2y+3x-2 is the same as 3x+2y-2.
2012-13: VCE at Parade College (Chemistry, English, Mathematical Methods, Physics and Specialist Mathematics).
2014-16: Bachelor of Science at La Trobe University (Mathematics and Statistics).
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2018-21: PhD at La Trobe University (Mathematics).

Only Cheating Yourself

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Re: VCE Methods Question Thread!
« Reply #3333 on: December 23, 2013, 07:16:34 pm »
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I still don't know where i am stuffing up!

Find the equation of the line containing each pair of points.  Put into ax+by+c=0

(2,-2) (0,1)

m=1--2=3  0-2=-2 M= -3/2

y-2=-3/2(x-2)

2y-4=-3x+6

3x+2y+10=0????????  thats the wrong answer where am i stuffing up?
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Conic

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Re: VCE Methods Question Thread!
« Reply #3334 on: December 23, 2013, 07:21:07 pm »
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I still don't know where i am stuffing up!

Find the equation of the line containing each pair of points.  Put into ax+by+c=0

(2,-2) (0,1)

m=1--2=3  0-2=-2 M= -3/2

y-2=-3/2(x-2)

2y-4=-3x+6

3x+2y+10=0????????  thats the wrong answer where am i stuffing up?

It should be y+2=-3/2(x-2), as you have y-(-2), which is y+2.
2012-13: VCE at Parade College (Chemistry, English, Mathematical Methods, Physics and Specialist Mathematics).
2014-16: Bachelor of Science at La Trobe University (Mathematics and Statistics).
2017-17: Bachelor of Science (Honours) at La Trobe University (Mathematics).
2018-21: PhD at La Trobe University (Mathematics).

Only Cheating Yourself

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Re: VCE Methods Question Thread!
« Reply #3335 on: December 23, 2013, 07:32:44 pm »
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It should be y+2=-3/2(x-2), as you have y-(-2), which is y+2.

yea my bad but I'm still making a mistake please help me

so yes it should be

y+2=-3/2(x-2)

2y+4=-3x+6
 
now here is where i think i stuff up
i bring the x to the left side to get rid of the -( as we want it in ax+by+c=0)
3x+2y=2
and then to bring the 2 to left side i becomes a positive so
3x+2y-2=0

Wholy crap, i spent like an hour thinking what i did wrong and i finally get it right, about god dam time !  thanks for your help guys
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Only Cheating Yourself

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Re: VCE Methods Question Thread!
« Reply #3336 on: December 23, 2013, 11:08:26 pm »
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Find the equation of the line that passes through the point of intersection of the lines whose equations are 7x-3y-19=0 and 3x+2y+5=0, given that the required line is parallel to the line with equations -5x-2y=3

i got c correct by m was wrong.  Here was my working out

I first used the elimination method to find x1 and y1

 7x-3y=19 *2
 3x+2y=-5 *3

14x-6y=38 (+)
9x+6=-15 (+)
=23x=23
x=1

y=7(1) - 3y=19
7-3y=19
-3y=12
y=-4

So far so good ?

now i need to find the gradient

-5x+2y=3
rearrange to make y the subject
2y=5x+3
y=5/2+3/2
M= 5/2

now i have my x1 and y2 (1,-4) and my gradient 5/2 i can solve this question
y+4=5/2(x-1)
2y+8=5x-5
2y=5x-13
5x-2y-13=0
y=-2=13

y=5/2x- 13/2

My answer is wrong.?
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Nato

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Re: VCE Methods Question Thread!
« Reply #3337 on: December 23, 2013, 11:36:46 pm »
+2
Find the equation of the line that passes through the point of intersection of the lines whose equations are 7x-3y-19=0 and 3x+2y+5=0, given that the required line is parallel to the line with equations -5x-2y=3

i got c correct by m was wrong.  Here was my working out

I first used the elimination method to find x1 and y1

 7x-3y=19 *2
 3x+2y=-5 *3

14x-6y=38 (+)
9x+6=-15 (+)
=23x=23
x=1

y=7(1) - 3y=19
7-3y=19
-3y=12
y=-4

So far so good ?

now i need to find the gradient

-5x+2y=3
rearrange to make y the subject
2y=5x+3
y=5/2+3/2
M= 5/2

now i have my x1 and y2 (1,-4) and my gradient 5/2 i can solve this question
y+4=5/2(x-1)
2y+8=5x-5
2y=5x-13
5x-2y-13=0
y=-2=13

y=5/2x- 13/2

My answer is wrong.?

Hey the main mistake i see is ou have copied down one of the equations wrong. In the question you had -5x-2y=3.
but when you were trying to find the gradient, you used -5x+2y=3
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Apink

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Re: VCE Methods Question Thread!
« Reply #3338 on: December 23, 2013, 11:55:14 pm »
+2
Find the equation of the line that passes through the point of intersection of the lines whose equations are 7x-3y-19=0 and 3x+2y+5=0, given that the required line is parallel to the line with equations -5x-2y=3

i got c correct by m was wrong.  Here was my working out

I first used the elimination method to find x1 and y1

 7x-3y=19 *2
 3x+2y=-5 *3

14x-6y=38 (+)
9x+6=-15 (+)
=23x=23
x=1

y=7(1) - 3y=19
7-3y=19
-3y=12
y=-4

So far so good ?

now i need to find the gradient

-5x+2y=3
rearrange to make y the subject
2y=5x+3
y=5/2+3/2
M= 5/2

now i have my x1 and y2 (1,-4) and my gradient 5/2 i can solve this question
y+4=5/2(x-1)
2y+8=5x-5
2y=5x-13
5x-2y-13=0
y=-2=13

y=5/2x- 13/2

My answer is wrong.?
Student0001 is right. You're on the right track though, give it another shot. Can you check the answer again, I think should be , not
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Only Cheating Yourself

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Re: VCE Methods Question Thread!
« Reply #3339 on: December 24, 2013, 12:02:49 am »
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Hey the main mistake i see is ou have copied down one of the equations wrong. In the question you had -5x-2y=3.
but when you were trying to find the gradient, you used -5x+2y=3

Thanks, these stupid errors are killing me! 
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Only Cheating Yourself

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Re: VCE Methods Question Thread!
« Reply #3340 on: December 24, 2013, 12:05:01 am »
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Student0001 is right. You're on the right track though, give it another shot. Can you check the answer again, I think should be , not

Yes, thanks both of use +.  Yes c does =-3/2

I don't want to sound like an idiot, but if I'm making these mistakes, do i really have a chance at methods?  I'm freaking out, i've nearly completed chapter 1 of the methods 1-2 concept book, i've done every question thus far obviously made mistakes but learnt from them.  I'm just really worried if its going to get hard and i get left behind.
'My belief is stronger than your doubt'

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Re: VCE Methods Question Thread!
« Reply #3341 on: December 24, 2013, 12:09:31 am »
+3
Don't stress on little things like that , we all make mistakes. The more questions you do, you will be more aware of the silly mistakes! Methods is hard but it's bearable if you keep on top.

Apink

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Re: VCE Methods Question Thread!
« Reply #3342 on: December 24, 2013, 12:41:52 am »
+4
Yes, thanks both of use +.  Yes c does =-3/2

I don't want to sound like an idiot, but if I'm making these mistakes, do i really have a chance at methods?  I'm freaking out, i've nearly completed chapter 1 of the methods 1-2 concept book, i've done every question thus far obviously made mistakes but learnt from them.  I'm just really worried if its going to get hard and i get left behind.

Don't freak out, I think you'll do well in Methods. Why? Because you seek help when you're having trouble and I think that's key in doing well in any subject. It's awesome that you're learning from your mistakes, that's what the questions are designed to do so don't let the number of mistakes bother you. Keep doing questions, keep asking us questions and keep learning from them.
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alchemy

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Re: VCE Methods Question Thread!
« Reply #3343 on: December 24, 2013, 09:24:49 am »
+1
Keep doing questions, keep asking us questions and keep learning from them.

This. Don't hesitate to PM me with any questions that you feel like your making a silly mistake on and don't want to post here.
And as long as you learn from your mistakes, everything is fine, you're on the right track  (:

Only Cheating Yourself

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Re: VCE Methods Question Thread!
« Reply #3344 on: December 24, 2013, 11:10:04 am »
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thanks guy, appreciate the help.  Having a good holidays! 

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