Simulation equations with a quadratic

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username4426374
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#1
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#1
2x^2-y^2=31
x+3y=1


So have I rearranged the second equation correctly? I have done: 3y=1-x
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RDKGames
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(Original post by Steve445)
2x^2-y^2=31
x+3y=1
Ok, so you know what the first step is?
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RDKGames
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(Original post by Steve445)
So have I rearranged the second equation correctly? I have done: 3y=1-x
You're not wrong, but you're also not quite there with this.

Whenever we solve two equations simultaneously, we are effectively looking for a solution(s) (x,y) which satisfy both equations at once. Graphically, these are the coordinates where the two equations intersect each other. At the point of intersection, the y-coordinates are the same, and the x-coordinates are the same.

Since we know that the y-coordinates are the same, this means that y in 2x^2-y^2=31 is exactly the same as the y in x+3y = 1. This means if you can rearrange the second equation for y on its own, then you can substitute that into the first equation and obtain a relation that x solely must satisfy.

So far, you have found what 3y is, but you need to know what y on its own is. So what's one more step??


Also, as a side note, you may be used to 'rearranging for y and subbing it into the other equation' type of approach, but I also said that the x-coordinates are the same. So there is nothing stopping you from rearranging the second equation for x instead and subbing that into the first equation. In fact, this is a much simpler alternative than that which you are doing because we avoid fractions this way and the algebra is less messy hence making less room for error.
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username4426374
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(Original post by RDKGames)
Ok, so you know what the first step is?
I know how to do the rest, but was my first bit correct

Edit: Nvm
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username4426374
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RDKGames
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(Original post by Steve445)
...
That's wrong.

You substituted in 3y = 1-x for y when you should've substituted in y = \dfrac{1-x}{3} instead.
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username4426374
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(Original post by RDKGames)
That's wrong.

You substituted in 3y = 1-x for y when you should've substituted in y = \dfrac{1-x}{3} instead.
Oh, okay... So what I did was completely wrong?
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RDKGames
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(Original post by Steve445)
Oh, okay... So what I did was completely wrong?
No, you just subbed in the wrong thing. The method is fine.
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idontkn
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Why not rearrange and get x to be the subject? And then plug that in to the quadratic.
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xAikx
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Would of been easier here to make x the subject of the second equation.

So x = 1 - 3y.

You can then substitute this expression for x in your first equation and solve for your possible values for x.

From there you can find y.
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username4426374
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So confusing man.
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username4426374
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#12
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How can stupid examiners expect you to do this on the new spec???
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username4426374
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#13
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Edexcel - "we are the most easiest exam board"
Me - Yeah, yeah, sure you are.
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username4426374
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And why is it that when I look online on how to do similar questioqu like this nothing comes up?
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xAikx
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(Original post by Steve445)
And why is it that when I look online on how to do similar questioqu like this nothing comes up?
This kind of simultaneous equation is actually civered in AS maths so maybe try looking at AS maths resources in this topic
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Muttley79
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(Original post by Steve445)
How can stupid examiners expect you to do this on the new spec???
This is not a new topic - it's been in the GCSE for ages, I would rearrange the second equation for x as it avoids fractions.
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