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May 15, 2026, 09:59:01 pm

Author Topic: BEC'S methods questions  (Read 119066 times)  Share 

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Re: BEC'S methods questions
« Reply #165 on: March 14, 2008, 02:17:49 am »
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Just remember the factor is always going to be divisor of the last term, since once you expand all the linear factors, the numbers multiply to give the last term, which i think is pretty neat.

bec

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Re: BEC'S methods questions
« Reply #166 on: March 14, 2008, 12:37:22 pm »
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thanks

what about this question - would i ever have to work it out by hand? if so, how - i can't get it!

Solve for x and y:
3x-7y + 6 = 0
x2 - y2 = 16

thank you!

bec

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Re: BEC'S methods questions
« Reply #167 on: March 14, 2008, 05:21:40 pm »
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hate to tell you this after all that working....but you read my question wrong!

Quote
Solve for x and y:
3x-7y + 6 = 0
x2 - y2 = 16

................so 3x 6+7y....

i tried to do it again but i've literally spent about 40min on this question and its just ridiculous, i keep making mistakes and getting different answers.
the book says:
so, if anyone feels up to tedious time-consuming work....

Collin Li

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Re: BEC'S methods questions
« Reply #168 on: March 14, 2008, 06:23:46 pm »
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Substituting this into yields:



Multiplying both sides by to kill the fraction:













Substitute these values of back into this equation found above: to find the corresponding and values:



Matching up with into coordinates gives us our solutions (in pairs): and
« Last Edit: March 14, 2008, 06:43:16 pm by coblin »

iamdan08

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Re: BEC'S methods questions
« Reply #169 on: March 14, 2008, 06:28:27 pm »
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I think you have transposed the original equation incorrectly
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Collin Li

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Re: BEC'S methods questions
« Reply #170 on: March 14, 2008, 06:39:52 pm »
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I think you have transposed the original equation incorrectly

Thanks, I corrected it!

bec

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Re: BEC'S methods questions
« Reply #171 on: March 15, 2008, 08:22:41 am »
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thanks coblin!

how about this one:
a question gave me the cubic equation y=ax3+bx2+cx+3 and a series of points so i could find the value of the constants a and b in terms of c (i got this part right)

   Then the next question was
Show that

um, how?

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Re: BEC'S methods questions
« Reply #172 on: March 15, 2008, 09:02:05 am »
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Always remember that Show means to prove that LHS = RHS or the first equation = second equation.

:)

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Collin Li

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Re: BEC'S methods questions
« Reply #173 on: March 15, 2008, 10:28:46 am »
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Always remember that Show means to prove that LHS = RHS or the first equation = second equation.

:)

Well, the idea is that you are not supposed to use information from both sides or both equations. For example, you shouldn't be merely verifying that , rather you should show you can get from the LHS to the RHS.

However, you definitely should have a look at both sides to get a hint as to how to manipulate one of the sides to get the other.

Since you are asked to show that , you have to start with what you know:



Looking at what we're supposed to show, there is a factor of out the front. Not a bad idea:



Now, there is a factor of in the equation we're supposed to show, so I'll take it out:



Now, to simplify the fraction, just use long division. I will not do the working here, but it should nicely fall out. In this case, I recommend using long division, to show the process. Synthetic division is dodgy because the method is not visible, and since you are already given the solution, it might look like you faked it.

bec

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Re: BEC'S methods questions
« Reply #174 on: March 15, 2008, 10:56:21 am »
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ok, that's very clear.

how about this one?

The graph of , intersects the line with equation in two points. The rightmost intersection has coordinates
a) Find the coordinates of the leftmost intersection
b) Find the exact value of b.

I figured it out but my method was dodgy and inefficient... how would you work it out?

Collin Li

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Re: BEC'S methods questions
« Reply #175 on: March 15, 2008, 11:21:14 am »
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What was your method? Sometimes a "dodgy and inefficient method" is just because you lack confidence in what was actually a method that demonstrated a good understanding of the material.

For example, the way I would have approached this (if it was tech free), was to have a look at the graph (without the dilation ), and then have a look at my first two intersections with :
.

The reason why I looked at the first two intersections is because these are the two intersections (as the question said) that the graph makes with , except they are squished or stretched by the dilation factor . We know that the rightmost intersection at is the second intersection, and this corresponds to the second intersection on the original graph .

Now, the exact value of falls out really easily now, because .

From this, it follows that the first (left side) intersection is

I guess a less confident student would think this method is clumsy, but it is more elegant than clumsy, in my opinion.

The reason why this works is because the way the question is worded, it starts looking at the intersections from the point . Since dilations sort of occur from as the centre, you can use this method. The situation becomes a bit more difficult if your domain does not start at zero, because now a dilation factor would affect how many solutions occur before or after the domain (so you can't simply assume that the first two intersections from the original graph correspond to the first two intersections on the dilated graph).
« Last Edit: March 15, 2008, 11:33:35 am by coblin »

bec

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Re: BEC'S methods questions
« Reply #176 on: March 15, 2008, 11:52:11 am »
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What was your method? Sometimes a "dodgy and inefficient method" is just because you lack confidence in what was actually a method that demonstrated a good understanding of the material.
well i think it was a combo of both- a bit of dodginess and also a lack of confidence
but you're right, i did use pretty much the same method as you did (but didn't express it particularly clearly in my working)

One more question: what does "collinear" mean? if it means "in a straight line", i don't know how to work out this question - could anyone help me?  (if it doesn't mean that, it'd be great if you could just tell me what it means and i'll try to work the q out myself)
Three cubic storage containers sit side by side with their visible faces in a vertical plane. The edge lengths of the boxes are m, x and n, where m<x<n. The top left corners of the 3 boxes are collinear.
a) If m=4 and x=6, find n
b) Express x in terms of m and n

(this is from p276 of mathsworld for anyone who has it - there are diagrams)

thankss

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Re: BEC'S methods questions
« Reply #177 on: March 15, 2008, 11:55:01 am »
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Yeah, that's what collinear means.
« Last Edit: March 15, 2008, 12:09:57 pm by coblin »

Collin Li

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Re: BEC'S methods questions
« Reply #178 on: March 15, 2008, 12:19:56 pm »
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The way I would do it is imagine the three boxes of length , and stacked from left-to-right (the exact order shouldn't matter). The order must be this, or the opposite (which is symmetrical), because (the middle-sized box) must be in the middle so that the boxes' top-left corners are collinear.

Now, to in order demonstrate collinearity, I will construct a coordinate system to do it easily: imagine there is a y-axis running down the left-side of the box, and an x-axis running along the bottom-side of the box.

The top-left coordinate of the first and second boxes are and , respectively. The top-left coordinate of the third box should be . If these three points are collinear, then a linear graph can be drawn through all these three points. This means that the gradient of should be the same as the gradient of .

The gradient of . The gradient of .





Note that whether these boxes are stacked left-to-right or right-to-left will not matter. The gradients you find might be negative, but when you equate them, the negatives will just cancel out!

For the second part, do the same thing I did, except without 4, and 6, just keep them as variables, and do the algebra to find . I've gotta go now!

Thanks to Ahmad and Neobeo for modelling this situation for me (they used similar triangles instead).
« Last Edit: March 15, 2008, 12:23:38 pm by coblin »

bec

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Re: BEC'S methods questions
« Reply #179 on: March 15, 2008, 12:23:06 pm »
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i figured out (a) but i still can't do (b) so i'll describe the diagram to you (i don't know how to copy in pictures!)

Quote
I don't see how a cubic container could have the dimensions  where . I think they mean rectangular prisms
Yes, they are cubes but it says in the question that we're only looking at "their visible faces in a vertical plane" - so basically it's a row of squares of increasing sizes:

There is a line with the equation
The first square has its top-left corner at (0,4), the second has its corner at (4,6) and the third at (10,9).

a) If m=4 and x=6, find n
n=9
b) Express x in terms of m and n

edit: sorry, too late! but you/neobeo/ahmad were right with the eqn!
« Last Edit: March 15, 2008, 12:25:11 pm by bec »