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Simultaneous Equation help?

I am trying to solve this simultaneous equation but I can't factorise it so either I am going wrong in the equation or I am missing out something while trying to factorise.

Anyway help will be appreciated.

3x^2 + y^2 = 28

3x + y = 8 (Linear)

8 - 3x = y

3x^2 + (8 - 3x)^2 = 28


3x^2 + 64 - 48x + 9x^2 = 28

12x^2 - 48x + 36 = 0
64 - 28 = 36.
Reply 2
Glutamic Acid
64 - 28 = 36.


My bad I meant to type 36 I still can't factorise the f*king thing should I be able to factorise that?
Reply 3
It does factorise, try dividing through by 12 and see if that makes it any clearer.
Reply 4
Thanks, got sorted now, it was a moment of not being able to see the wood for the tree's.
Reply 5
Hit another problem on simultaneous question:
Is this working out correct?
2x + y = 8 Y=2x - 8
4x^2 + 3y^2 = 52

4x^2 + 3(2x - 8)^2 = 52

4x^2 + 3(4x^2 - 32x + 64) = 52

4x^2 + 12x^2 - 96x + 192 = 52

16x^2 - 96x + 140 = 0
Reply 6
Assuming the first line should be 2x + y = 8 and rearrangement, the rearrangement should actually be y = 8 - 2x

Edit: Having said that, when substituted in, it squares to the same solution, so your final answer is, as far as I can see, correct.
Reply 7
Shakaa
Hit another problem on simultaneous question:
Is this working out correct?
2x + y = 8 Y=2x - 8
4x^2 + 3y^2 = 52

4x^2 + 3(2x - 8)^2 = 52

4x^2 + 3(4x^2 - 32x + 64) = 52

4x^2 + 12x^2 - 96x + 192 = 52

16x^2 - 96x + 140 = 0

Method is fine, but you've got the first line wrong. Just complete the square and back substitute and you're done.:smile:
Reply 8
I realise this is often frowned upon in these forums, but can someone show me the steps in order to factorise/complete the square of the following in order the reach the answer? Its really doing my head in :/

2x + y = 8 Y=2x - 8
4x^2 + 3y^2 = 52

4x^2 + 3(2x - 8)^2 = 52

4x^2 + 3(4x^2 - 32x + 64) = 52

4x^2 + 12x^2 - 96x + 192 = 52

16x^2 - 96x + 140 = 0
Shakaa

16x^2 - 96x + 140 = 0


Last post for the day; and on this occasion only:

Divide through by 16:

x^2-6x + 35/4=0

Complete the square:

(x-3)^2 -9 +35/4 = 0

(x-3)^2 = 1/4

and I'm sure you can finish that.
Reply 10
ghostwalker
Last post for the day; and on this occasion only:

Divide through by 16:

x^2-6x + 35/4=0

Complete the square:

(x-3)^2 -9 +35/4 = 0

(x-3)^2 = 1/4

and I'm sure you can finish that.


I can't figure out why -9 +35/4 transforms into 1/4 :confused:
Shakaa
I can't figure out why -9 +35/4 transforms into 1/4 :confused:


(X-3)^2 -9 + 35/4 = 0 --> (X-3)^2 = 9 - 35/4

convert 9 to it's fractional equivalent ( 9 = 9/1)
so 9 = 36/4

substituting in

(X-3)^2 = 36/4 - 35/4

leaving

(X-3)^2 = 1/4

Hope that makes it more clear :smile:
Reply 12
GingerGoat
(X-3)^2 -9 + 35/4 = 0 --> (X-3)^2 = 9 - 35/4

convert 9 to it's fractional equivalent ( 9 = 9/1)
so 9 = 36/4

substituting in

(X-3)^2 = 36/4 - 35/4

leaving

(X-3)^2 = 1/4

Hope that makes it more clear :smile:



I don't full understand how it works but I understand what's happened now thanks.

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