[blueray]

I have done part i) And have the correct answer of
Pos a
x^2=y^2+4

Pos b

(x+0.95)^2 = (y+1.05)^2 + 4

Edit (thanks guys)

I don't know how to do the LAST part, and would be grateful, if you could go through it

Thank you!

Question bellow.

[notnek]

I'm a bit confused by your question. Which part are you asking about?

Could you maybe ask your question again in a different way?

[Intriguing Alias]

(Original post by blueray)
I have done part i) And have the correct answer of
Pos a
x^2=y^2+4

Pos b

(x+0.95)^2 = (y+1.05)^2 + 4

I was wondering why would the real answer not be x= and y=
Why is it like the above answer, if the question says write down 2 equations in X an d Y?

I also don't know how to do the second part, and would be grateful, if you could go through it

Thank you!

Question bellow.

If I understand correctly; you're asking why it's okay to leave those equations in that form?

If so; it's simply that they didn't ask you to make y or x the subject (subject means make it x= ... or y= ... ); so as long as you have some equation relating x and y you've done what they've asked.

For the hence show; multiply the brackets in the second equation out and see if that helps you see what to do.

[notnek]

OK I think I now understand your question

It says, write down two equations "in x and y". An equation "in x and y" doesn't have to look like y=... or x=..., it just has to be an equation that includes the variables x and y.

If the question said, "write an equation with x/y as the subject" or "write an equation for x/y in terms of y/x" then you would be required to write equations of the form x/y=...

[blueray]

(Original post by hassi94)
If I understand correctly; you're asking why it's okay to leave those equations in that form?

If so; it's simply that they didn't ask you to make y or x the subject (subject means make it x= ... or y= ... ); so as long as you have some equation relating x and y you've done what they've asked.

For the hence show; multiply the brackets in the second equation out and see if that helps you see what to do.

(Original post by notnek)
OK I think I now understand your question

It says, write down two equations "in x and y". An equation "in x and y" doesn't have to look like y=... or x=..., it just has to be an equation that includes the variables x and y.

If the question said, "write an equation with x/y as the subject" or "write an equation for x/y in terms of y/x" then you would be required to write equations of the form x/y=...

Oh ok cheers guys Can you help me do the next part?

Yes I multiplied it out, but the sums wont add up to get that answer
Could you start me off and i'll do the rest? Or go through it? THanks.

[CharlieBoardman]

For part ii) expand your brackets for pos B to get
Re arrange to

For pos a, we can re arrange to have

So we can now solve these simultaneously:
Re arrange to get your

[Aeonstorm]

The first part is like that because it means that the equation must have both x and y in it. If you really want to change x^2 = y^2 + 4 into y = sqrt(x^2-4) for example, you could, but it is somewhat pointless.

For part (ii), try solving your two equations simultaneously. I recommend expanding your Pos b equation and then subbing in (y^2+4) for x^2. Things will just cancel out.

Part (iii) is just a continuation of this - you need to square both sides and then sub in (x^2 - 4) for y^2. Tbh, part (ii) is actually somewhat redundant and just a stepping stone I guess to help you get to part (iii); you could get to the quadratic in one go.

[blueray]

(Original post by CharlieBoardman)
For part ii) expand your brackets for pos B to get
Re arrange to

For pos a, we can re arrange to have

So we can now solve these simultaneously:
Re arrange to get your

Oh ok thanks for that! I was just struggling to find what to sub I get it now! I'll remember to look out to sub numbers with letters

[k9markiii]

(Original post by blueray)
I have done part i) And have the correct answer of
Pos a
x^2=y^2+4

Pos b

(x+0.95)^2 = (y+1.05)^2 + 4

Edit (thanks guys)

I don't know how to do the second part, and would be grateful, if you could go through it

Thank you!

Question bellow.

You have to continue to expand your answer for B and then combine A and B into a single equation so that you cancel out the squared terms. Then rearrange that equation so it equals y in terms of x and substitute into equation A or B. Good luck. I'm not going to go through exactly how it's done.

[CharlieBoardman]

(Original post by blueray)
Oh ok thanks for that! I was just struggling to find what to sub I get it now! I'll remember to look out to sub numbers with letters

Ha no worries

What are the answers on the mark scheme for iii)? I just want to see if I'm heading in the right direction.

[blueray]

(Original post by notnek)
a

(Original post by hassi94)
y.

(Original post by Aeonstorm)

Part (iii) is just a continuation of this - you need to square both sides and then sub in (x^2 - 4) for y^2. Tbh, part (ii) is actually somewhat redundant and just a stepping stone I guess to help you get to part (iii); you could get to the quadratic in one go.

I was wondering, why would you have to square both sides? Then sub?? Please explain why you said this.
Thanks!

(Original post by CharlieBoardman)
Ha no worries

What are the answers on the mark scheme for iii)? I just want to see if I'm heading in the right direction.

Can all of the above show me how to get there?

Final answer
x=4.25
y= 3.75

[GreenLantern1]

(Original post by blueray)
I was wondering, why would you have to square both sides? Then sub?? Please explain why you said this.
Thanks!



Can all of the above show me how to get there?

Final answer
x=4.25
y= 3.75

Use the quadratic formula to get x. Then sub x back into your original equation and you have got y.

[Aeonstorm]

When i said square both sides, i meant squaring both sides of 2.1y = 1.9x - 0.2, so that you have 4.41y^2 which you can replace with 4.41(x^2-4). Alternatively, you can just solve to get y = (1.9x-0.2)/2.1, and then square this and sub into x^2 = y^2 + 4. Essentially its the same thing either way.

[blueray]

(Original post by GreenLantern1)
Use the quadratic formula to get x. Then sub x back into your original equation and you have got y.

Oh so I have to use quadratic equation on this?

x^2 +1.9x+0.9025

Because the question is asking me to form a quad equation in x ???

[blueray]

(Original post by Aeonstorm)
When i said square both sides, i meant squaring both sides of 2.1y = 1.9x - 0.2, so that you have 4.41y^2 which you can replace with 4.41(x^2-4). Alternatively, you can just solve to get y = (1.9x-0.2)/2.1, and then square this and sub into x^2 = y^2 + 4. Essentially its the same thing either way.

If you said this, then why would green say the bellow??
And you also haven't said WHY you square both sides? What in the question or what do you think in your head, that tells you to "square both sides"?

Once I get this then I will understand it fully

(Original post by GreenLantern1)
Use the quadratic formula to get x. Then sub x back into your original equation and you have got y.

[Aeonstorm]

(Original post by blueray)
Oh so I have to use quadratic equation on this?

x^2 +1.9x+0.9025

Because the question is asking me to form a quad equation in x ???

You do want a quadratic equation, but thats not it. Just take y = (1.9x - 0.2)/2.1 and substitute into x^2 = y^2 + 4

[Aeonstorm]

(Original post by blueray)
If you said this, then why would green say the bellow??
And you also haven't said WHY you square both sides? What in the question or what do you think in your head, that tells you to "square both sides"?

Once I get this then I will understand it fully

Because green has already gotten x, before he tells you how to get y. I'm talking about how to get the quadratic in x in the firs tplace.

As for you you need to square, look at my above post. If y = (1.9x - 0.2)/2.1, and you want to sub it into x^2 = y^2 + 4, you have to square both sides of y = (1.9x - 0.2)/2.1 to get y^2 = (1.9x - 0.2)^2/4.41, before you can sub it in.

If you just push that through, you should get a quadratic of 0.8x^2 + 0.76x - 17.68 = 0
Then just use the quadratic formula to get x, which is 4.25
Then subs x = 4.25 into x^2 = y^2 + 4 to get y

[CharlieBoardman]

(Original post by blueray)
I was wondering, why would you have to square both sides? Then sub?? Please explain why you said this.
Thanks!

Can all of the above show me how to get there?

Final answer
x=4.25
y= 3.75

So you square both sides of 2.1y=1.9x-0.2.
This gives you
We know that so we sub that in to get:

Rearrange to get:
Use quadratic formula for your values of x. Sub your values of x into one of the equations to get your values of y

[CharlieBoardman]

(Original post by blueray)
If you said this, then why would green say the bellow??
And you also haven't said WHY you square both sides? What in the question or what do you think in your head, that tells you to "square both sides"?

Once I get this then I will understand it fully

You needed to square it so that it could be turned into a quadratic equation when you subbed in your y^2=x^2-4.

EDIT: The question doesn't tell you to square both sides, you just have to see that it has to be done, in order to turn in into a quadratic

[blueray]

(Original post by CharlieBoardman)
So you square both sides of 2.1y=1.9x-0.2.
This gives you
We know that so we sub that in to get:

Rearrange to get:
Use quadratic formula for your values of x. Sub your values of x into one of the equations to get your values of y

Cheers I get it now!