#1

Yay, i come here to ask the pit for help. I know there are some smarty pants lurking in the shadows.

so here it is! (^ means squared here)

x^ + y^ = 16

y = 3x - 1

I got as far as

x^+ (3x-1)^ = 16

x^ + 9x^ - 6x +1 = 16

10x^-6x - 15 = 0

(something x^ -/+ ) (something x +/- )

^^^^^ Factorise? I can't work out what numbers to put in. So i tried the forumla

-b +/- √b^-4ac

----------------------

2a

and suprise suprise got nowhere apart from....

+/- 4.81 (2.d.p)

+/- 7.19 (2.d.p)

???????? Yeah, i have a whole 4 pages of **** like this (all involving simultaneous equations) so get ready for cooly fun time.

Any help is greatly appreciated thanks.

so here it is! (^ means squared here)

x^ + y^ = 16

y = 3x - 1

I got as far as

x^+ (3x-1)^ = 16

x^ + 9x^ - 6x +1 = 16

10x^-6x - 15 = 0

(something x^ -/+ ) (something x +/- )

^^^^^ Factorise? I can't work out what numbers to put in. So i tried the forumla

-b +/- √b^-4ac

----------------------

2a

and suprise suprise got nowhere apart from....

+/- 4.81 (2.d.p)

+/- 7.19 (2.d.p)

???????? Yeah, i have a whole 4 pages of **** like this (all involving simultaneous equations) so get ready for cooly fun time.

Any help is greatly appreciated thanks.

#2

Im doing that in maths atm. Unfortunately im incredibly thick. Sowwie.

#3

lol, i can do t3h easy ones. But what the hell is that. Its only the numbers in the factorising that i can't work out. And i don't know if the forumla answers are right, or where to put the two numbers it gives. :S

#4

Leave in the surd form I think??? It is better than giving decimals/

#5

The answer is Pi

It's always Pi

It's always Pi

#6

The answer is Pi

It's always Pi

haha, I wish.

I'm doing Math's at Uni Level now and it's ridiculously hard.

I'll put some of my questions up soon

EDIT:

{xEZ | x2<16} = {-3,-2,-1,0,1,2,3} = {xEZ | |x|<4}}

Another EDIT:

{x,yEZ|-2<x<3^0<y<xo(x,y)} = ?

*Last edited by Kingyem0c0re at Oct 27, 2007,*

#7

i'm pretty sure your numbers in the (5x-na)(3x+nb) factorization are going to be decimals.

#8

The last equation doesn't factor, and I got 1.561 and -0.961 for X. (My calculator doesn't lie )

So that makes Y=3.863, -3.883

Its just one math problem, if its a real pain, move on.

So that makes Y=3.863, -3.883

Its just one math problem, if its a real pain, move on.

#9

x^2 + (3x-1)^2 = 16

x^2 + 9x^2 - 6x + 1 = 16

10x^2 - 6x - 15 = 0

now plug these coefficients into the quadratic equation and you should get the answers from the last post: a = 10, b = -6, c = -15.

x = [-(-6) + sqrt(6^2 - 4*10*(-15))]/(2*10) = 1.56095

x = [-(-6) - sqrt(6^2 - 4*10*(-15))]/(2*10) = -0.960952

I'm sure you can get the y-values from here!

x^2 + 9x^2 - 6x + 1 = 16

10x^2 - 6x - 15 = 0

now plug these coefficients into the quadratic equation and you should get the answers from the last post: a = 10, b = -6, c = -15.

x = [-(-6) + sqrt(6^2 - 4*10*(-15))]/(2*10) = 1.56095

x = [-(-6) - sqrt(6^2 - 4*10*(-15))]/(2*10) = -0.960952

I'm sure you can get the y-values from here!

*Last edited by bentfocus at Oct 27, 2007,*

#10

why cant you just take the square root of x^2 + (3x-1)^2 = 16?

x + (3x - 1) = 4

4x= 5

x = 5/4

of course, that may not be how youre supposed to do it. youre using a technique i did 2 years ago and i dont use in calculus..

x + (3x - 1) = 4

4x= 5

x = 5/4

of course, that may not be how youre supposed to do it. youre using a technique i did 2 years ago and i dont use in calculus..

#11

why cant you just take the square root of x^2 + (3x-1)^2 = 16?

x + (3x - 1) = 4

4x= 5

x = 5/4

.

Well, the square root of x^2 + (3x-1)^2 = 3.16228*sqrt(x^2 - 0.6x + 0.1),

not: x + (3x - 1).

If you did that it's like saying sqrt(4 + 4) = 2 + 2 = 4, when it is clearly

sqrt(4 + 4) = sqrt(8) = sqrt(2*4) = 2*sqrt(2)

#12

Wait lol i just realised i put the numbers into the forumla wrong.

(-b) would be (+6) because the b is -6 already.

I eventually, have done the whole paper. Thanks guys!

(-b) would be (+6) because the b is -6 already.

I eventually, have done the whole paper. Thanks guys!

#13

doing maths GCSE in year 10... a year early, and doing english lit, language, french and RE in year 10 too...

#14

the answer is 5

#15

Completing the square is so much easier.

10x^2 - 6x - 15 = 0

x^2 - 6/10x - 15/10 = 0

(x - 3/10)^2 - 9/100 - 15/10 = 0

(x - 3/10)^2 = 159/100

x - 3/10 = +/-root(159/100)

x = 3/10+/-root(159/100)

10x^2 - 6x - 15 = 0

x^2 - 6/10x - 15/10 = 0

(x - 3/10)^2 - 9/100 - 15/10 = 0

(x - 3/10)^2 = 159/100

x - 3/10 = +/-root(159/100)

x = 3/10+/-root(159/100)

#16

Hope that helps.

#17

I might sound like an idiot but can you not just do a square root job on the top equation so you get x+y=4?.....No..?

EDIT: Yeah, the guy above me is right, substitution should work

EDIT: Yeah, the guy above me is right, substitution should work

*Last edited by con job at Oct 28, 2007,*

#18

^^^we have a winner

except b = -6 not 6

btw thats not hard

try uni maths mate you'll cry

except b = -6 not 6

btw thats not hard

try uni maths mate you'll cry

*Last edited by henza_x at Oct 28, 2007,*

#19

I might sound like an idiot but can you not just do a square root job on the top equation so you get x+y=4?.....No..?

EDIT: Yeah, the guy above me is right, substitution should work

3^2 + 4^2 = 5^2

3 + 4 =/= 5

#20

The answer is Pi

It's always Pi

wrong, its 42

#21

3^2 + 4^2 = 5^2

3 + 4 =/= 5

Probably should have checked did my theory make sense before I said it

#22

Yay, i come here to ask the pit for help. I know there are some smarty pants lurking in the shadows.

so here it is! (^ means squared here)

x^ + y^ = 16

y = 3x - 1

How old are you? Cause we saw this when I was 14 or 15, it's easy.

If you're not sure about it, check with your calculator (Texas if you got one)

#23

I might sound like an idiot but can you not just do a square root job on the top equation so you get x+y=4?.....No..?

dont think so... when you route them all, the 2 can be positive or negative