Sorry! It's about maths again, I need some more help! Watch

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CHAD
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Hi
Sorry to bug you again. It's about simualtaneous equations again. The question is:

Solve the simualtaneous equations

x^2 - 4y^2 = 24
x + 2y = 6

When I posted this b4 I didnt understand how number like 36 and 12y^2 were obtained, so could you explain please?

BTW, I know the answers are
x = 5
y = 0.5

I just dont understand how you get there.

Sorry to ask again!

Thank-you!
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Darkness
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(Original post by CHAD)
Hi
Sorry to bug you again. It's about simualtaneous equations again. The question is:

Solve the simualtaneous equations

x^2 - 4y^2 = 24
x + 2y = 6

When I posted this b4 I didnt understand how number like 36 and 12y^2 were obtained, so could you explain please?

BTW, I know the answers are
x = 5
y = 0.5

I just dont understand how you get there.

Sorry to ask again!

Thank-you!
Read my solution on the second page of your other post. If things still aren't clear, ask again for help.
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Bhaal85
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(Original post by CHAD)
Hi
Sorry to bug you again. It's about simualtaneous equations again. The question is:

Solve the simualtaneous equations

x^2 - 4y^2 = 24
x + 2y = 6

When I posted this b4 I didnt understand how number like 36 and 12y^2 were obtained, so could you explain please?

BTW, I know the answers are
x = 5
y = 0.5

I just dont understand how you get there.

Sorry to ask again!

Thank-you!
You know that x+2y=6 therefor x=6-2y sub into x^2-4y^2=24 to get

(6-2y)^2-4y^2=24

expand to get:

36-24y+4y^2-4y^2-24=0

12-24Y=0 therefore Y=0.5, sub Y=0.5 into x+2y=6 therefore X=5.
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SasunD
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(1) x^2 - 4y^2 = 24
(2) x + 2y = 6

x = 6 - 2y

so into equation (1)

(6 - 2y)^2 - 4y^2 = 24
(36 - 24y +4y^2) - 4y^2 = 24

get together all the terms

36 - 24y = 24 ==>(the +4y^2 and -4y^2 cancel out)

so -24y = 24 - 36 = -12
so 24y = 12
therefore y 0.5

put y = 0.5 into (2) x + (2*0.5) = 6
x + 1 = 6
so x = 5

x = 5
y = 0.5
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Bhaal85
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Are you sure you get simultaneous equations that are quadratic? I thought at GCSE level they where linear equations?
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username9816
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(Original post by Bhaal85)
Are you sure you get simultaneous equations that are quadratic? I thought at GCSE level they where linear equations?
These Simultaneous Equations are AS Level Pure 1..
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Bhaal85
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(Original post by bono)
These Simultaneous Equations are AS Level Pure 1..
That is what I thought, but the thread starter said they had a GCSE mock, surely this wont be on it?
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username9816
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(Original post by Bhaal85)
That is what I thought, but the thread starter said they had a GCSE mock, surely this wont be on it?
linear simultaneous equations ddint even turn up on my gcse exam....but yes, these harder quad. ones r AS Level pure 1
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Bhaal85
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(Original post by bono)
linear simultaneous equations ddint even turn up on my gcse exam....but yes, these harder quad. ones r AS Level pure 1
Its not hard.
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username9816
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(Original post by Bhaal85)
Its not hard.
well obviously, i meant "harder than linear which is extremely easy"
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4Ed
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(Original post by bono)
well obviously, i meant "harder than linear which is extremely easy"
this is not of pure 1 standard since the quadratics cancel out.... so it's still linear in a sense.... all you do is substitute in and square out, which i'm sure GCSE maths pupils are capable of doing... the quadratics are just there to scare people off.
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username9816
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(Original post by 4Ed)
this is not of pure 1 standard since the quadratics cancel out.... so it's still linear in a sense.... all you do is substitute in and square out, which i'm sure GCSE maths pupils are capable of doing... the quadratics are just there to scare people off.
lol - maths does well in scaring people.
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Darkness
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(Original post by bono)
These Simultaneous Equations are AS Level Pure 1..
No they're not because the squared terms cancel.
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username9816
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(Original post by Darkness)
No they're not because the squared terms cancel.
i wasnt really looking, sorry.

yes, they do cancel out........

but if it came up on P1 Exam, i wouldnt mind!!
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Darkness
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(Original post by bono)
i wasnt really looking, sorry.

yes, they do cancel out........

but if it came up on P1 Exam, i wouldnt mind!!
lol. I'm sure you wouldn't!
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Bhaal85
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(Original post by bono)
i wasnt really looking, sorry.

yes, they do cancel out........

but if it came up on P1 Exam, i wouldnt mind!!
The only topic I have ever found slighlty hard at P1 was Transformations, and Transformations of trig functions. Pain.
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username9816
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(Original post by Bhaal85)
The only topic I have ever found slighlty hard at P1 was Transformations, and Transformations of trig functions. Pain.
i hate transofrmations, not hard, but its just tedious...
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Bhaal85
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(Original post by bono)
i hate transofrmations, not hard, but its just tedious...
Ok simple question:

Y=x^3 is transformed firstly by a stretch on the x-axis by S.F 3 and translated by

(0)
(-3)

What is the result of the transformation?
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byb3
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(Original post by Bhaal85)
Ok simple question:

Y=x^3 is transformed firstly by a stretch on the x-axis by S.F 3 and translated by

(0)
(-3)

What is the result of the transformation?
If I remember correctly...

Stretch on the x-axis by a factor of 3, means that:

(1,1) ==> (3, 1)

So x ==> x/3

Translated:

x stays the same
y ==> y + 3

y + 3 = (x/3)^3
y = (x/3)^3 - 3

Was I close?
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Bhaal85
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(Original post by byb3)
If I remember correctly...

Stretch on the x-axis by a factor of 3, means that:

(1,1) ==> (3, 1)

So x ==> x/3

Translated:

x stays the same
y ==> y + 3

y + 3 = (x/3)^3
y = (x/3)^3 - 3

Was I close?
Yep, you where right. I didn't realise you could tackle it that way. I just did it this way:

(1/3 times X)^3 - 3 = (X/3)^3 - 3
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