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## Topic review (newest first)

mathsyperson
2006-01-10 06:01:50

For future reference, if you want to ask for help, then create a new topic rather than posting on someone else's.

Anyway, for these simultaneous equations you need to scale them up so that they have a common term, then combine them to give a simple linear equation.

2x-4y=16 --> (multiply by 3) 6x-12y=48
3x+6y=-12 --> (multiply by 2) 6x+12y=-24

Add these equations together: 6x-12y+6x+12y=48-24
12x=24
x=2

Knowing this, you can easily work out the other value.

2*2-4y=16
-4y=12
y=-3

Check: 3*2+6y=-12
6y=-18
y=-3

It works! So your final answer is x=2, y=-3

angela
2006-01-10 04:55:59

2x-4y=16
3x+6y=-12

help

kylekatarn
2005-09-15 01:21:22

you're welcome :)

lkomarci
2005-09-14 19:20:17

Excellent indeed
and once more...the day is saved by kylekatarn. THANK YOU

kylekatarn
2005-09-14 07:49:35

#### MathsIsFun wrote:

Excellent!

thanks:)

MathsIsFun
2005-09-14 07:21:18

Excellent!

kylekatarn
2005-09-14 06:53:57

#### Code:

```let z = 2-5x

dz
—— = -5
dx

dz = -5.dx

⌠    3dx       3 ⌠    dx        3 1 ⌠   5dx        3 ⌠ dz
│——————————— = -.│————————— = - —.—.│————————— =- ——.│————
│        4/5   4 │      4/5     4 5 │      4/5    20 │ 4/5
⌡ 4(2-5x)        ⌡(2-5x)            ⌡(2-5x)          ⌡z

3
let α = - ——  and β = 4/5 to make things more clear.
20
┌      ┐
│ -β+1 │
⌠ dz    ⌠ 1         ⌠ -β       │z     │         α   -β+1
α│——— = α│——— dz = α ⌡z   dz = α│——————│ + C = ————.z     + C
│  β    │  β                   │ -β+1 │       -β+1
⌡ z     ⌡ z                    └      ┘

-β+1 = 1/5

α      3
———— = - —
-β+1     4

putting everything together we get:

3      1/5
- —(2-5x)    + C
4```

and that's it : )

kylekatarn
2005-09-14 06:42:44

The solution is correct.
I'm writting the explanation rigth now

lkomarci
2005-09-13 02:33:28

hey kyle...i got another one for you. i want u to solve this one, because i think that the solution in the book is not correct.  the book was written by some of my teachers, they probably wrote the darn book in one night. so many mistakes.

ok here it goes:

∫ 3dx / 4(2-5x)^4/5 =

the solution given in the book goes like this:

I=-3(2-5x)^1/5  /  4 + C

please...

kylekatarn
2005-09-12 09:49:32

ok! I'm always here for integration problems : )
Calculus IS my favorite subject!

lkomarci
2005-09-12 06:03:18

hey thanx for your detailed reply kylekatarn.
i dunno why i didn't think of writing that root as en exponent. i do that a lot

thanx again. i'll write back if i got some more problems

kylekatarn
2005-09-12 00:10:45

∫ [ (3x-5)^(1/4) ]^3 dx = ∫ (3x-5)^(3/4) dx
z=3x-4
dz/dx=3
dz=3.dx

∫ (3x-5)^(3/4) dx = (3/3) ∫ (3x-5)^(3/4) dx = (1/3) ∫ 3 (3x-5)^(3/4) dx = (1/3) ∫ (3x-5)^(3/4) 3.dx =
= (1/3) ∫ z^(3/4) dz = (1/3) [ z^(3/4+1) ] / (3/4+1) +C = (1/3)(4/7) [ z^(7/4) ] + C = (4/21) (3x-5)^(7/4) + C
and that's the answer.

kylekatarn
2005-09-11 23:55:42

If you are learning integration you are less n00b than you think...: )

lkomarci
2005-09-11 23:20:48

hi everyone, plz don't laugh at me, i'm a math noob, i'm currently learning integrals and i got a problem. i dont understand how to solve this exercise:

∫ 4th sqrt(3x-5)³ dx    (it's not 4 x, it's 4th root)

1.  now...i usually take this under the brackets and derivate it:
3x - 5=t
3dx=dt /:3
dx=dt/3

the solution should be:   I = 4 X 4th sqrt (3x-5)^7  / 21 + C

my math-english is not that good so.. this 4th sqrt should mean fourth square root

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