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Simultanous equations

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F257

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It seems somone on this server is very fond of procreational activites.I owuld have posted something in dediction but not allowed to post pornographic materail instead so ...

 

 

 

 

 

Simultaneous Equations (Linear)

Recreated from a Revision Book. One presumes there is no copyright on Maths

 

Professor%20John%20Frink.gif

 

1. Let us assume that you are asked to solve the following pair of equations

 

2x = 6 - 4y

-3 - 3y = 4x

 

We must rearrange both equations such that

 

ax + by = c

 

where a,b, and c are numbers

 

Thus

 

2. 2x + 4y = 6

-4x -3y =3

 

(rember you are potentialy swapping numbers from both sides of the equation, so they may become negative and vice versa)

 

 

Be sure to label those kwrazy equations!

 

2x + 4y = 6 (1)

-4x -3y =3 (2)

 

 

3. You must 'match' the coefficents of either the X's OR Y 's in both equations . This is done by multiplying either one or both of the equations by a suitable number. A postive may match a negative (its far less problematic than rhesus's anyway)

 

Therfore if we multiply

 

 

2x + 4y = 6 (1)

 

 

by 2 we should thus get

 

4x + 8y = 12 (1)

 

and you'll notice the 4 of the coefficent of X matches

 

-4x -3y =3 (2)

 

(Negative no problem)

 

Be sure to renumber those kwrazy equations !

 

4x + 8y = 12 (3)

-4x -3y =3 (4)

 

3. If the Coefficents are the same (both +ve or both -ve) then SUBTRACT

If the Coefficents are different (one +ve and one -ve) then ADD

 

The latter rule applies here and thus

 

(3) + (4) 0x +5y =15

 

Solve this remaining equation

 

5y =15 - - > y= 3

 

 

4. Now subsitue this value back into equation (1)

 

2x + 4 x 3 = 6 - - > 2x + 12 = 6 - - > 2x = -6 - - > x = -3

 

5. Now subsitute both equations back into equation (2) and it should work . If not

 

you-have-failed-please-die.jpg

 

(Na just try again or ask someone else)

 

(-4 x -3) -( 3 x 3) = 3

x = -3 , y = 3

 

 

 

Re-edited for clearing confusion.Danke Phantasm :)

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lol no i just had spare time to waste before whatever i wanted to watch on telly come :P Which part you lost on? I actually manged to learn this myself from some website a long time ago. Havent done it in ages but can work it out

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lol wait you can use this

 

http://www.hellam.net/algebra/simul1.html

 

its great :)

 

Ok look at

 

http://www.bbc.co.uk/schools/gcsebitesize/maths/algebra/simultaneoushirev1.shtml

 

_______________________________________________________________________________________

Its probaley clearer on the link but....

 

Solve these simultaneous equations and find the values of x and y.

 

* Equation 1: 2x + y = 7

* Equation 2: 3x - y = 8

 

As you can see they are already in the format ax +by = c

 

And also the Y coefficent is already matched up. Since the two coeffiecentsa are different you need to add

 

Add the two equations to eliminate the ys:

 

* 2x + y = 7

* 3x - y = 8

* ------------

* 5x = 15

* x = 3

 

 

Now you got x =3, put into equation 1 (as i state in the guide above, not in the link. its the same results either way but i like to double check)

 

Equation 1 ) 2x + y = 7

 

2 x 3 + y = 7

 

6 + y = 7

y=1

 

Equation 2 ) 3x - y = 8

 

3 x 3 - 1 = 8

 

It works so

 

X = 3, Y =1

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Scroll to the bottom an you 'll see they havent arranged

 

Equation 1: y - 2x = 1

 

Equation 2: 2y - 3x = 5

 

Into the ax +by = c format but you can

 

Equation 1: -2x +y = 1

 

Equation 2: -3x +2y = 5

 

Now if we multipoly equation 1 by 2 to make coefficent of Y to match up

 

Equation 1: -4x +2y = 2

 

Equation 2: -3x +2y = 5

 

This comes to

 

 

-x = -3

 

 

means

 

x =3

 

aubstitue into equation 1

 

Equation 1: 2 x 3 +y = 1

 

means

 

6 + y = 1

 

rearange to

 

6 + 1 = y

 

y = 7

 

 

Equation 2: -3x +2y = 5

 

to check this is correct put both y and x into equation 2

 

 

Equation 2: -3 x 2 +2 x 7 = 5

 

x = 3 , y =7 . same results as the link

 

(lol sorry ran out of the number of quotes im allowed :P)

 

Lol i was only ever good on this part of the course ^^ Hopeless at the graphs/quadractic's

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OMFG .... I always hate math... blehhh

 

I loved maths so much i took sn A lelvel in it even though i got all Us in my exams :P

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Oops Sorry i thought I'm still in f|a forum for games... OK got it so my browser didn't clean old forum cache I visited last time by accident and talking about math :huh:

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ROFL F25, i was sittin in math today, totally stuck on this, and i whipped out my laptop and started browsing the forums. Stumbled upon this gem, and now i understand!!!

 

God bless you =F|A= Priest :thanks

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Oops Sorry i thought I'm still in f|a forum for games... OK got it so my browser didn't clean old forum cache I visited last time by accident and talking about math :huh:

 

lol well...maths can be fun :lol: And rofl val. Hope this works - dont blame me if it dosent, blame the BBC :P

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O_O what the hell xD this is so simple solved in an other way xD

 

2x = 6 - 4y

-3 - 3y = 4x

 

take first equation and fint the x --> x = 3 - 2y

 

then substitute in the other equation --> - 3 - 3y = 12 - 8y

 

solve --> 5y = 15 --> y = 3

 

substitute in first equation --> 2x = -6 --> x = -3

 

 

 

and that's done :D

 

 

i use that method only when i 've got to do whith harder equations, like in analytic geometry when u gotta find common points between two circumferences or parables like this:

 

x^2 + y^2 + 5x + 6y + 13

x^2 + y^2 + 12x + 7y + 2

 

in that way u can get rid of squared terms and find x & y :D

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O_O what the hell xD this is so simple solved in an other way xD

 

2x = 6 - 4y

-3 - 3y = 4x

 

take first equation and fint the x --> x = 3 - 2y

 

then substitute in the other equation --> - 3 - 3y = 12 - 8y

 

solve --> 5y = 15 --> y = 3

 

substitute in first equation --> 2x = -6 --> x = -3

 

 

 

and that's done :D

 

i like being awkwared :D

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