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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!

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!

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!

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.

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!

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.

(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

(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

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..

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

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.

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!!

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...

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?

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?

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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