[gavinlee]

Hi,

OK, so I thought I'd mastered simultaneous equations a long time ago, but now under Core 1 revision I've come across one which I just can't get the right answer for. Can anyone tell me what I have done wrong?

x+4(y-1)=3(x-3)+2y [a]
3(x-1)=5(x-1) [b]

rearranging both a & b:
2x-2y-5=0 [a]
2x+3y-5=0 [b]

b - a.
5y-10=0
5y=10
y=2.

if y=2,
2x-2y-5=0 [a]
2x-4-5=0
2x=9
x=4.5

So I get x=4.5 & y=2. The book has x=2.5 & y=0, which do fit in if you work it out, but I thought my method was correct. What is it that I've done wrong?

[olipal]

[starkrush]

I don't see how you re-arrange:

to
,

where are the y terms?

[aeyurttaser13]

as olipal said, the equation should be 5y = 0 not 5y - 10 = 0

[gavinlee]

OK, thanks for looking through it and spotting it for me

[GreenLantern1]

3(x-1)=5(x-1) is incorrect. If you divide by x-1 you have 3=5 which is clearly an impossibility!

[olipal]

(Original post by GreenLantern1)
3(x-1)=5(x-1) is incorrect. If you divide by x-1 you have 3=5 which is clearly an impossibility!

What about if ?

In this instance though, the OP has written down the original question wrong (though the reduced equations a and b yield the correct answer).

[GreenLantern1]

(Original post by olipal)
What about if ?

In this instance though, the OP has written down the original question wrong (though the reduced equations a and b yield the correct answer).

Fair enough.

[elldeegee]

(Original post by GreenLantern1)
3(x-1)=5(x-1) is incorrect. If you divide by x-1 you have 3=5 which is clearly an impossibility!

I thought this!

(Original post by olipal)
What about if ?

In this instance though, the OP has written down the original question wrong (though the reduced equations a and b yield the correct answer).

But then i also thought this... (I'm sounding like an Aldi advert now :P)

to be able to solve 3(x-1)=5(x-1) would you HAVE to use trial-and-error (i say trial-and-error, but really plugging 1 into it isn't very taxing)?

I thought that the first method would have been justified if it hadn't given out a contradiction, or is there some sort of trick that i am not seeing? like the whole "prove 2=1" story, where it turns out that a=b=0 and so you cannot divide by it, or something....

[olipal]

(Original post by elldeegee)
to be able to solve 3(x-1)=5(x-1) would you HAVE to use trial-and-error (i say trial-and-error, but really plugging 1 into it isn't very taxing)?

Trial and error? Surely just expanding out and looking at what you've got would suffice:









Sorry if I've missed the point of your post!

[elldeegee]

(Original post by olipal)
Trial and error? Surely just expanding out and looking at what you've got would suffice:




Sorry if I've missed the point of your post!

haha it's that youve missed my point but also stupidity on my part.

Why i said trial and error i don't really know, as expanding would be the best bet.

I cant really rephrase my question properly, so ignore it now :P

[olipal]

(Original post by elldeegee)
haha it's that youve missed my point but also stupidity on my part.

Why i said trial and error i don't really know, as expanding would be the best bet.

I cant really rephrase my question properly, so ignore it now :P

Well in general the rule is you can't divide by 0, so you can only confidently divide by something if you know for sure it can't be 0. Not only can it lead to contradictions, but you can also miss solutions to equations ( for instance).

[elldeegee]

(Original post by olipal)
Well in general the rule is you can't divide by 0, so you can only confidently divide by something if you know for sure it can't be 0. Not only can it lead to contradictions, but you can also miss solutions to equations ( for instance).

oh duh!
so as you cannot divide through by ?

[olipal]

(Original post by elldeegee)
oh duh!
so as you cannot divide through by ?

You got it! Just be wary when dividing by anything

[elldeegee]

(Original post by olipal)
You got it! Just be wary when dividing by anything

hehe thanks!
I will be forever on the lookout now