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# Simultaneous Equation help? watch

1. 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
2. 64 - 28 = 36.
3. (Original post by 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?
4. It does factorise, try dividing through by 12 and see if that makes it any clearer.
5. Thanks, got sorted now, it was a moment of not being able to see the wood for the tree's.
6. 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
7. 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.
8. (Original post by 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.
9. 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
10. (Original post by 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.
11. (Original post by 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
12. (Original post by Shakaa)
I can't figure out why -9 +35/4 transforms into 1/4
(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
13. (Original post by 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

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

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Updated: February 10, 2010
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