₦airaland Forum

[quote author=awesome_benji]awesome_benji: @OP especially, simultaneous eqn in 3unknowns.
Given that:
a1x + b1y + c1z + d1 = 0
a2x + b2y + c2z + d2 = 0
a3x + b3y + c3z + d3 = 0
find x, y & z and leave d answer in matrix form. Please, please & please, help me 2 solve dis as d
lecturer has promised 2 bring it in d exam. Thanks.[/quote][a1 b1 c1][x]=[-d1]
[a2 b2 c2][x]=[-d2]
[a3 b3 c3][x]=[-d3]

| a1 b1 c1 : -d1|
| a2 b2 c2 : -d2|
| a3 b3 c3 : -d3|

Now u can solve ur three variables by manipulating matrix(adding and subtracting lines)
e.g 2x+y-3z=-4
4x-2y+z=9
3x+5y-2z=5
Solution: substituting data in ur matrix

| 2 1 -3 : -4|
| 4 -2 1 : 9 |
| 3 5 -2 : 5 |

For row(2): add -2 times row(1) plus 1 times row(2)
For row(3): add -3 times row(1) plus 2 times row(3)

| 2 1 -3 : -4 |
| 0 -4 7 : 17|
| 0 7 5 : 22|

Now,
For row(3): add 7 times row(2) plus 4 times row(3)

| 2 1 -3 : -4 |
| 0 -4 7 : 17|
| 0 0 5 :207|

Back to algibraic eq.

2x+y-3z=-4
-4y+7z=17
69z=207
If u fiddle with it now u should get
x=2
y=1
z=3

Krucifax: Okay here's one for the board. Solve below using equations of absolute value


a) | x | = 5

b) | x - 2 | = 7

c) | 89 x + 4 | = -2

d) | -11 x - 9 | = 0

e) 2 | x + 5 | = 10

f) 3 | -9 x - 7 | - 2 = 13

g) -5 | -x + 2 | + 8 = -12

h) | -3 x - 2 | / 3 = 7

i) | (x - 5) / 4 | = 6

j) | x + 1 | = |-4 x + 3|

k) | x + 3 | = - |-x - 3 |

I am going to answer a few selected question, not easy to type this stuff on a phone.

(a) |x|=5

x=5 or -x=5
Two possible cases as absolute values are either positive or negative

x=5 or x=-5
(b) |x-2|=7

x-2=7 or -(x-2)=7
x=9 or x-2=-7
x=9 or x=-5
(c) |89x+4|=-2
No need solving ..

(d) |-11x-9|=0
-11x-9=0
x=-9/11
Now this big motherfu.cker
(j) |x+1|=|-4x+3|

x+1=|-4x+3|

x+1=-4x+3 or x+1=-(-4x+3)
5x=2 or x+1=4x-3
x=2/5 or 3x=4
x=2/5 or x=4/3

Ok, others are not as bad azz as this one^^^
NB: I left school with an excruciating 2 class lower.

mouyoursawre: Can u do me a favour by helping me to find the square root of -3.

Let's try
Assuming x^2=-3

X=sqrt(-3)=(-3)^0.5

X=(3 * -1)^0.5

X=(3)^0.5 * (-1)^0.5

X=1.73205 * i

X=1.73205i
A complex number

Obinoscopy: Hi youngsage,

Nice try! U got the answer right! But I'm sorry, your step is just SO WRONG!

First of, your derivation of eqn iii from eqn ii is wrong. This because log(A+B) is never equal to logA+logB

Also your eqn v is wrong as well. xlogx+ylogy cannot be added directly cos they've got no common factor. Thus it would be wrong to just multiply x with y and logx with logy to give xylogxy

Finally the answers x and y are pure integers and not decimals as u got. You only had to approximate cos u knew it will lead u to d answer u so desire

I'm sorry I had to criticise u work. Nice attempt though.

Try this: x^x+y^y=46912, x+y=2

Lol, no answer for dat, if u plot both equations the never meet each other. In other words u fabricated it

Truman155: man ps can u use ya calculator and tell me what u will get as values of a,b and c if u change the last equation to a^3+b^3+c^3=32 ps just the answers

a=2-2i
b=2+2i
c=4

shady26: ositadinma1,billionare=same person.
shreld_b,regynus=same person

Hahahaha, I can remember the last time I laughed this hard. I have never separated from my body. Not my thing, besides I am insomniac

^^^ lol, what type of equipment are u using, whatever its I know it must be really expensive, 3A men, that's something!!!

Fhemmmy: That is really nice, Sir . . But the 40K could be a deposit and by the time the phone lands in Nigeria, you would be able to get the balance put together . . Looking forward to your business

Will it come with accessories?

Fhemmmy: Happy New year to you and we could you one MINT SII at a cost of only 52K Naira . . . Looking forward to doing business with you soon

Mr Fhemmy, I only have 40k ooo, can it do?

calculusx: [size=30pt]Hi all, a friend of mine went for an interview sometime ago and was asked to pronounce this simple equation mathematically. e^0 = 1. Can anyone in the house pronounce this mathematically?[/size]

Exponential of zero is equal to one

Funny thing is anything raised to power zero gives u one, osita raised to 0 is one, I am a little tipsy

ositadima1: x+cosx=2.154

x+cosx=2.154+0.i

x+[e^(-ix)]/2+[e^(ix)]/2=2.154

x=3.15392 (appr.)

converted cosx to its hyperbolic form

lol, ok I cheated....

The realistic way to solve it is to plot y=x+cosx against y=2.154, if u are good at graphs u should get 3.1 something at there intrsection, mind u cosx is in radians.

I am quoting my self, not good ooo.
If u plot y=2.154 against y=x+cosx u get x=3.1 something pending on how nice u layed ur grid, again x is in radians.

Anyway, let's c what others will come up with

Richiez: x(x^2+ 4)= 0

what x(x^2+ 4)= 0 means is that (x+0)(x^2+ 4)= 0
therefore one of the x is x=0

now x[x^3/24- x/2 + 1] = 1.154 is a different situation.

actually: [x^3/24- x/2 + 1] = 1.154/x.

Richiez: if you check the R.H.S, u'l find out that the value of x cannot exceed 2.154, and this is a small number compared to the next term which is to be divided by 6!, so this makes other terms quite insignificant.
and for the factorization, if you have an eqn, say, x(x^2+ 4)= 0 , one of the values of x must be 0. same goes for x[x^3/24- x/2 + 1] = 1.154, one of the values of x shud be 1.154
okay lets check
x+cosx=2.154
when x=1.154, x+cosx= 1.154 + cos(1.154) = 1.154 + 0.999 = 2.154 (approx)

Now, u kind of cheated, because maclaurin's series is based on radians (x is in radians). but when u fed back the answer u calculated, u did ur cos in degrees. I am coming...

^^^cool

Richiez: SOLUTION
x+cosx=2.154.....(1)
we have to apply maclaurin's series for cosx= 1- x^2/2! + x^4/4!......
hence, x+ 1- x^2/2! + x^4/4!......=2.154
x^4/24- x^2/2 + x =2.154-1
x^4/24- x^2/2 + x =1.154
by factorization; x[x^3/24- x/2 + 1] = 1.154
therefore; x= 1.154 or x^3/24- x/2 + 1 = 1.154

with this we have succeeded in getting on value of x=1.154, solve for other values

I disagree with u man, u choose the first three terms of d maclaurin's series, if my memory still dey, u can only make this assumption if u sure x<1, in dat case the rest of d series becomes insignificant. In this case u looking for the x, besides even ur factorization is kinda off. Ok, try feeding ur answer into the equation and c if it adds up.

Time2talk: Try solve this:
(X^2+Y^2)dx + (2XY)dy = 0

ODE, hmmm?

y=c-[x(X^2+Y^2)]/2XY
Where c is d arbitrary constant.
Do u want to c d steps?

lordZOUGA: do you know that you can browse the internet from the command line? Yet, people use browsers..

I like ur app. nah, it's a piece of ingenuity, besides GUI makes things easier for the less geeky.

^^^ dat is d first derivative

Odehfamily: Open the bracket,by multiplying log by x and y. U wil hav logx+logy=log5

U cant multiply log, its neither a number nor a variable.

Instead: log(xy) = logx + logy

Beretta92: no, log(xy)=log x + log y

He still got a correct answer,

Odehfamily: Excellent

Is log(x+y)=log5 same as logx+logy=log5

dabrake: nooooh!


Though, the question isn't wrong. I proved it wasn't but to find that a, b or c na die. Hopefully, the answer will appear before the day runs out.

Guy dont stress ur self, I used a programming calc. and I got this answer;
a=4.87119
b=1.5644-2.51229i
c=1.5644+2.51229i

This is one set of answers, they are like 3 set total and they are all complex numbers, I doubt anybody can solve this equations by hand.

If a+b+c=8, a^2+b^2+c^2=16 and a^3+ b^3+c^3=64 find a,b, and c

Proflakes: x+cosx=2.154, find x

x+cosx=2.154

x+cosx=2.154+0.i

x+[e^(-ix)]/2+[e^(ix)]/2=2.154

x=3.15392 (appr.)

converted cosx to its hyperbolic form

lol, ok I cheated....

The realistic way to solve it is to plot y=x+cosx against y=2.154, if u are good at graphs u should get 3.1 something at there intrsection, mind u cosx is in radians.

Mr Reyginus

.

[quote author=chreld_b]I dont really get where you are going with your question about straw man fallacy. But as far as I'm concerned any person's opinion relating to this topic devoid of logic is a matter of fact a straw man.[/quote]Ok

Tolzeal: it a fogd questn I think...am n ma 5th page solvn stil no ansa...

Men, we have three variables and third power here, this ish is too complex to be solved by hand

dabrake: i continued from here, solved another 2 pages
.
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And I arrived at equation 1(a + b + c) = 8.


E neva happen before

Hahahaha, what d dude did was wrong sha

(a+b+c) ^2 = 64 this makes no sense

double post

BrutusOj: Waoo..nice tread.my gurus in d house,kindly make t d subject of d formular in d expresion below" s=ut+1/2at^2" I.e d second equation of motion(s=ut+half atsquare).am waitin

To solve this one I am going to use the completing the square method...

s=ut+(0.5)at^2
Multiply by 2/a
2s/a=(2u/a)t+t^2

t^2+(2u/a)t+....=2s/a+....
Now, to make our equation a perfect square we add (u/a)^2 to both sides.

t^2+(2u/a)t+(u/a)^2=2s/a+(u/a)^2
Lets factor this baby...

(t+u/a)^2=2s/a+(u/a)^2

t+u/a=(2s/a+(u/a)^2)^0.5

note its plus minus, meaning u have two answers

t=(2s/a+(u/a)^2)^0.5-u/a

t=-(2s/a+(u/a)^2)^0.5-u/a

BrutusOj: Waoo..nice tread.my gurus in d house,kindly make t d subject of d formular in d expresion below" s=ut+1/2at^2" I.e d second equation of motion(s=ut+half atsquare).am waitin

To solve this one I am going to use the completing the square method...