From my experiance I can see from this equation that in can't be factorised unless you use imaganary numbers which is too high a level for this. Are you sure you typed the eqautions in properly?

Reply 3

username3446470

OP

20

Original post by PandahatDude

First you rearange equation 1 by subtracting

2y

from both sides to get the equation

x=13-2y

Then you substitute this value into equation 2 to get

(13-2y)^2+y^2=9

which you the expand to

169-52y+4y^2+y^2=9

This is simplified to

5y^2-52y+160=0

From my experiance I can see from this equation that in can't be factorised unless you use imaganary numbers which is too high a level for this. Are you sure you typed the eqautions in properly?

yeah sorry should be for the seconds equation x squared minus y squared equals nine

Reply 4

PandahatDude

6

Original post by jonjoshelvey21

yeah sorry should be for the seconds equation x squared minus y squared equals nine

Okay, so you would instead end up with

3y^2-52y+160=0

This can then be put into the quadratic formula to find

y=\frac{40}{3}

or

y=4

You then substitute these values into the equation

x=13-2y

to find that when

y=\frac{40}{3},x=-\frac{41}{3}

and when

y=4,x=5

I hope this helped

Reply 5

username3446470

OP

20

Original post by PandahatDude

Okay, so you would instead end up with

3y^2-52y+160=0

This can then be put into the quadratic formula to find

y=\frac{40}{3}

or

y=4

You then substitute these values into the equation

x=13-2y

to find that when

y=\frac{40}{3},x=-\frac{41}{3}

and when

y=4,x=5

I hope this helped

thanks it did a lot. btw what are imaginary numbers ?

Reply 6

PandahatDude

6

Original post by jonjoshelvey21

thanks it did a lot. btw what are imaginary numbers ?

If you ever take further maths in A Level then you get what is known as imaginary numbers. It is used when you are trying to square root a negative number which is impossible as any number times by itself is alway positive (e.g.

-1^2=1

) and therefore this is a problem when trying to solve complex eqautions which need you to find the sqaure root of a negative.

We say that the

\sqrt{-1}=i

. This then leads to lots of other crazy things where you can plot imaginary numbers alongside "real" numbers on a graph etc. I won't go into to much detail but if you want try putting the

\sqrt{-1}

in your calculator and it will come up with a math error as the number does not exist.

Reply 7

username3446470

OP

20

Original post by PandahatDude

If you ever take further maths in A Level then you get what is known as imaginary numbers. It is used when you are trying to square root a negative number which is impossible as any number times by itself is alway positive (e.g.

-1^2=1

) and therefore this is a problem when trying to solve complex eqautions which need you to find the sqaure root of a negative.

We say that the

\sqrt{-1}=i

. This then leads to lots of other crazy things where you can plot imaginary numbers alongside "real" numbers on a graph etc. I won't go into to much detail but if you want try putting the