Simultaneous equations

[kezia_88]

Sorry I'm not sure which catogory this goes under! But i'm doing my gcses and was wondering if anyone could help me in doing math:

For simultaneous equations why do you have to multiply some of them and how do you know what by please help I really need it

thanks

[Peter Pedant]

Hi kezia ! (your name is one of the sisters of Job isnt it? Jemima, kezia and someone else...)

OK I think you should probably go back to your maths master with that. I think the question is OK and not stoopid.

Simultaneous eqns......if you multiply one by 4 all the way through then the answer x =     and y = doesnt change.

So you multiply one of the equations so that the Xs get to be the same.....

Like 2x + 3y = 5

and 4x + 5y = 9,

then multiply the first one by two so you get 4x in each equation

Now you only have 4 x in each equation you can take one away from the other and yippee you are only left with an equation with y in it.....

 

Geddit? I hope this helps.....

 

PP

[Tim Baxter]

If it doesn't then you have to work on either equation until you get 2 terms identical.

a, 3x + 4y = 32

b. 6x + 5y = 48

What's the easiest way to get 2 terms identical?

it must be to multiply 'a' by 2 on each side, therefore,

6x + 8y = 64, now that you've got identical terms in equation (6x) you can subtract 'b' from modified 'a'

which gives you 6x-6x = 0, 8y-5y = 3y, 64-48 = 16.

Because the sum of the 'x' subtraction has resulted in zero (remember 0x = 0) you are left with 0x + 3y = 16, so, divide both sides by 3 and you have y = 16/3 which is 5 1/3 or 5.3 recurring.

To go back to the original pair you could multiply 'a' by 5 and 'b' by 4 which would give you 'y' as common terms (20y in each) then you would do the same subtraction again to reduce one term to zero etc.  You could do similar operations to the number terms to make them equal but it's the easiest way that you're looking for every time.

Good Luck

[jake-the-peg]

With a lot of mathematics there is more than one way to get an answer.

Sometimes schools teach a particular method and then you hear a different method from somebody who was at school twenty years before and it gets mixed up in your mind and it doesn't help.

Having said that with simple sets like in Peters example I like to take one equation and manipulate it until it is of the form X= something or Y= something

so 2x+3y=5  (subtract 3y from both sides)

2x=5 - 3y (divide by 2)

x= 2.5 -1.5y

Then replace all the x values in the the other equation with this

4 (2.5 -1.5y) + 5y = 9

or 10 - 6y + 5y = 9

or 10 - y = 9 (add y to both sides)

10=9+y => 10 - 9 = y

y=1

put this in to the equation of your choice

4x + 5y=9

4x+ 5 = 9

4x=9 - 5

4x = 4

x=1

check in the other equation

2x+3y=5

yup that works

It's longer but it always felt more comfortable to me

PS If the above confuses you in any way please feel free to ignore it immediately

[Peter Pedant]

all we now is for someone to say,

actually it is easier with matrices.....

and it will blow Kezia's head off

[kezia_88]

Lol yea sounds scary i have my exam on tues arghhh! btw I dont have a sis on here peter sowwi :(

[Tim Baxter]

You're quite right Peter, you've just taken me back 40 years to when my school year was being 'experimented on' with matrices, Venn diagrams and the like.

Good Luck Kezia

[fo3nix]

I realise this has already been answered fairly substancially, but thought i'd give my view of it anyway.
------------------------------------------------------

(1) 2x + 3y = 5

(2) 4x + 5y = 9

OK so first, either the x's or the y's have to be the same. Right now they are different (2 is not the same as 4, 3 isn't the same as 5). So make them the same by multiplying one of the equations.

Let's multiply: (1) x2, and call it (3)

(3) 4x + 6y = 10

so now we have (3) and (2):

(3) 4x + 6y = 10

(2) 4x + 5y = 9

now, i've done the multiplication to get one of the numbers equal. here i've made the x's equal. so now we focus on this, and say to ourselves "same sign substract". this means that if the two x values (the 4) have the same sign, then we subtract the two equations, (3)-(2). if the signs are different, say if the first was -4 and the other was +4, then we add the two. since here the signs are the same, we subtract ("same sign subtract", remember).

(3)-(2): we do this for each side of the equation (each side of the equals sign):

(4x + 6y) - (4x + 5y) = 10 - 9
therefore, y = 1

this is because the x's cancel (4x-4x=0), and the rest is simple.

so now we go back to any of our equations, (1), (2) or (3), and put in y=1. let's try it for (1):

(1) 2x + 3y = 5

as i said, put in y=1:

(1) 2x + 3 = 5

2x = 5 - 3
2x = 2
x = 1

there we have it. y=1, x=1. let's make sure this is true in (2):

(2) 4x + 5y = 9
if x=1 and y=1, we say:
4 + 5 = 9

which is very true! so we are correct. problem solved.

[fo3nix]

if you decide to do this at a-level or degree level, you will be shown a technique known as Gaussian elimination, which does essentially the same thing but with more than just two equations at the same time! cool eh :-)

[Tim Baxter]

Is this the Gauss who invented magnetism?  I geuss not!

[Tim Baxter]

Additionally, is there a process to solve trinomial equations?

[fo3nix]

i'm not familiar with trinomial equations.. what are they?

and yes, this is the same Gauss as Gauss' law in electromagnetism, gauss' theorem/divergence theorem from vector calculus etc.

Gauss did not invent magnetism though. but he certainly did a lot :)

[Tim Baxter]

Forgiven my pun? I geuss so!

[jean88]

isnt it suppose to be helping the kid? Dont confuse her! I agree with peter I do mine like his. I dont know is the old fashion way or not but all different schools teaches different ways. Some of the new ways are too confusing. My kid was confused with the new ways, now I teach her myself, my way, and she is a A* student. Who cares old fashion or not, it works and is easy to understand. Talking about the standards of teaching now-a-days...

[fo3nix]

exactly right jean88, regardless of whether the method is old or not, if it works and is easy to understand and remember, then who cares?

tim baxter: i've just got what you meant. trinomial equations as in 3 equations to solve 3 unknowns, etc. yes, you can solve this with Gaussian elimination. you can even do it for 10 unknowns and 10 equations :)