Simultaneous Help (1 Viewer)

[Lukybear]

From Cambridge:

Consider the equN 12x^2-4xy+11y^2=64 & 16x^2-9xy+11y^2=78
a) By letting y=mx show that 7m^2+12m-4=0
b) Hence solve the two equations simultaneously

Any help would be great on booth a and b, since i couldnt get any.

EDIT: Opps sorry changed
please help

 

[lyounamu]

From Cambridge:

Consider the equN 12x^2-4xy+11y^2=62 & 16x^2-9xy=11y^2=78
a) By letting y=mx show that 7m^2+12m-4=0
b) Hence solve the two equations simultaneously

Any help would be great on booth a and b, since i couldnt get any.

Um...

 

[Lukybear]

bump?

 

[lyounamu]

bump?

No your question semes little bit flawed..

EDIT: Ah, it's fixed now. Okie dokie

 

[Lukybear]

sorry, i didnt noe how to highlight the edit...

 

[georgefren]

Hmm I'm not sure if its possible.

 

[Lukybear]

Its from cambrdige 3unit yr 11 so....

+im sure lyounamu will figure it out

 

[georgefren]

I got

16 + 5xy + -4x² = 0

if thats any help?</pre>

 

[lyounamu]

I think your quesition is still little bit flawed...

I can get that into

that form x3m^2 + x2m + x1 = 0

but not into what you asked there...

let me try again.

I am getting:
88m^2/7 + 123m/7 - 4 = 0

 

[Lukybear]

how does it solve??

its nt in form of ax^2+bx+c=0

 

[Lukybear]

opps the first equN is 12x^2-4xy+11y^2=64
instead at first i wrote =62, if that helps...

cambridge might be flawed... u wanna know the answer? it might help... but this is fully beyond me

 

[georgefren]

Still don't get it. I'm on

4x^2-5mx^2-22m^2x^2-14=0

but that doesnt help.

 

[lyounamu]

opps the first equN is 12x^2-4xy+11y^2=64
instead at first i wrote =62, if that helps...

cambridge might be flawed... u wanna know the answer? it might help... but this is fully beyond me

I got it now.

wait, for the solution

12x^2 - 4xy + 11y^2 = 62
16x^2 - 9xy + 11y^2 = 78

so if you sub y=mx

you get:
12x^2 - 4mx^2 + 11m^2x^2 = 64
16x^2 - 9mx^2 + 11m^2x^2 = 78

Now you equate the 64 and 78 because you want to get RID OF THEM.

So by equating you get:
936x^2 - 312mx^2 + 858m^2x^2 = 4992 ...(1)
1024x^2 - 576mx^2 + 704m^2x^2 = 4992 ...(2)
Then (1)-(2):

so you get -88x^2 + 264mx^2 + 154m^2x^2 = 0
divide everything by 22x^2
you get -4 + 12m + 7m^2 = 0

done!

 

[Lukybear]

Answer is x=1, y=-2 or x=-1, y=2 or x=7/3 y=2/3 or x=-7/3, y=-2/3

 

[Lukybear]

I got it now.

wait, for the solution

cmon tell us alrdy

 

[lyounamu]

cmon tell us alrdy

It does take a fair bit of working out..

Yeah I got that equation.

Now just solve for m from that equation using quadratic formula.

Then sub that in and work out x and y value.

Hope that helped.

 

[Lukybear]

I got it now.

wait, for the solution

12x^2 - 4xy + 11y^2 = 62
16x^2 - 9xy + 11y^2 = 78

so if you sub y=mx

you get:
12x^2 - 4mx^2 + 11m^2x^2 = 64
16x^2 - 9mx^2 + 11m^2x^2 = 78

Now you equate the 64 and 78 because you want to get RID OF THEM.

So by equating you get:
936x^2 - 312mx^2 + 858m^2x^2 = 4992 ...(1)
1024x^2 - 576mx^2 + 704m^2x^2 = 4992 ...(2)
Then (1)-(2):

so you get -88x^2 + 264mx^2 + 154m^2x^2 = 0
divide everything by 22x^2
you get -4 + 12m + 7m^2 = 0

done!

wow nice work
thxs so much...

also nice try georgefren

 

[lyounamu]

wow nice work
thxs so much...

So, for the rests, I am sure you will get them since you will have the values of m using the quad formula. Or do you still need help with them?

 

[Lukybear]

im good
thxs

 

[lyounamu]

Ah, I am feeling generous today.

m = -2 or 2/7 using quad formula:

sub them into either (1) or (2)

so you have (for example from (1)):

when m = -2
12x^2 - 4 . -2 . x^2 + 11 . 4 . x^2 = 64
64x^2 = 64
x^2 = 64/64
x = 1 or -1

so since y = mx, y = -2 or 2

Try the same method of m=2/7

good luck.