Clarification on simultaneous equations. (1 Viewer)

[atBondi]

Ok so i have 4 equations, with 4 variables, if i got all the terms down to
say for example...
A= xB
A= xC
A= xD

where x is any random number,
for some reason when i sub them back into one of the equations i get the wrong answer, but if i sub it into the only equation i havent used it works..

So does his mean, that i had to make use of all 4 equations and i could only get the answer by subbing them into the equation i havent used?

If yous dont understand what im saying i can post my working, but its a fairly simple question and the working is long, so i havent typed it up.
thanks.

 

[adomad]

A= xB
A= xC
A= xD
this means that b=c=d? if you have 4 unknowns, you need 4 equations. i tinhk that if u post the question up we could help you

 

[emmcyclopedia]

Yeah, I think I understand where you're going.

Pretty much, when you have 3 eq'ns to solve simultaneously, you can start with the first 2. You'll solve it for a particular variable (e.g. x), then put your answer for x into the 3rd remaining equation. You can't put it back into the equations you've just used, or you'll end up only showing that the LHS is the same as the RHS.

Post up your question, you'll get more clarification then

 

[atBondi]

ok so the 4 equations are:
3a+4b+ 4c=0 (1)
27a+12b + 4c = 0 (2)
a+2b+4c+ 8d = 8 (3)
27a + 18b + 12c + 8d = -8 (4)

Then i went (2) - (1)
and got b= -3a (5)

Then (5) into (1)
9a/ 4 (6)

Then (5), (6) into (3)
d= (2-a) / 2 (7)

And this is where my problem arises,
When i sub 5,6,7 into (3), my a=0 which is wrong
But when i sub 5,6,7 into (4), i get the correct answer...a=4
So my question was pretty simple and i think emmcy was right, so pretty much i cant sub it back into the equation i just used..and i have to sub it into (4) which i havent used..?
Or have i made so heaps stupid mistake and now i look like a fool?

 

[emmcyclopedia]

yeah, so you solved for d in terms of a by subbing all values into eqn 3.

so, when you get b, c and d, you would put them all into eqn 4 - you've already used eq'n 3, thus you'd end up with 0 = 0 if you did it that way.

 

[sexisash]

Here's how it works if you have 4 equations 1, 2, 3 and 4:
Use equations 1 and 2 to make equation 5 (3 unknowns)
Use equations 3 and 4 to make equation 6 (the SAME 3 unknowns)
Use either 1 and 3 or 2 and 4 to make equation 7 (the SAME 3 unknowns).

Now you have 3 equations 5, 6 and 7.
Use 5 and 6 to make 8 (2 unknowns)
Use 5 and 7 or 6 and 7 to make 9 (the SAME 2 unknowns).

Now you have 2 equations. Solve them simultaneously to find 2 of your unkowns.

To find the other one, substitute the ones you've found into any of the equations with 3 unknowns. Then substitute that one into the one with 4 unknowns (since you've found 3).

Hope this helps,

Sasha