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

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.

Thank-you!
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.

Thank-you!

Read my solution on the second page of your other post. If things still aren't clear, ask again for help.
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.

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.
(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
Are you sure you get simultaneous equations that are quadratic? I thought at GCSE level they where linear equations?
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..
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?
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
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.
Bhaal85
Its not hard.

well obviously, i meant "harder than linear which is extremely easy"
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.
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.
bono
These Simultaneous Equations are AS Level Pure 1..

No they're not because the squared terms cancel.
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!!
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!
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.
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...
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?
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?
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