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andrewburton
07-07-2011, 05:27 PM
Guys, please help with this complex numbers simultaneous equations question, as I keep geting different answers. If i times equation 1 by (2+j) then, eq 2 by (eq (3-j). it should be the same for when i do the some calculation to b respectivly. But it is not.

Have a go:

2=(3-j)a-(5-j2)b
12=(2+j)a+(1+j6)b

Thank you

Andy

tkhunny
07-07-2011, 06:33 PM
First, "times" is nto a verb.

Second, what does it mean?

"j2" Is that j^2 = -1?
"j6" Is that (j^4)(j^2) = 1*(-1) = -1?

mmm4444bot
07-07-2011, 06:41 PM
complex numbers simultaneous equations

2=(3-j)a-(5-j2)b

12=(2+j)a+(1+j6)b

This notation is not standard.



Does j = sqrt(-1) ?

Does j2 mean j^2 and j6 mean j^6 ?

If the answer to each of these questions is yes, then the system simplifies to:

eqn1: 2 = (3 - j)a - 6b

eqn2: 12 = (2 + j)a





If i times equation 1 by (2+j) then, eq 2 by (eq (3-j). it should be the same for when i do the some calculation to b respectivly.

Can you please rephrase these statements ? I'm having some difficulty understanding them.

The expression (eq (3-j) has mismatched parentheses, for one thing. I'm also not sure why you typed "eq" in that expression or what you mean by "do the same calculation to b". What calculation? You've described at least two.

soroban
07-07-2011, 10:07 PM
Hello, Andy!

\text{I have seen }j = \sqrt{\text{-}1}
\text{And I have seen }2+ 3i\text{ written }2 + i2

Your game plan is correct.
You must be making algebraic errors.
. . (There are plenty of opportunities!)


\text{Solve: }\:\begin{array}{cccc}(3-j)a-(5-2j)b &=& 2 & [1] \\ (2+j)a+(1+6j)b &=& 12 & [2] \end{array}
\begin{array}{cccccc}\text{Multiply [1] by }(1+6j)\!: & (1+6j)(3-j)a - (1+6j)(5-2j)b &=& 2(1+6j) & [3] \\ \text{Multiply [2] by }(5-2j)\!: & (5-2j)(2+j)a + (1+6j)(5-2j)b &=& 12(5-2j) & [4] \end{array}

\text{Add [3] and [4]: }\:(1+6j)(3-j)a + (5-2j)(2+j)a \:=\:2(1+6j) + 12(5-2j)

. . . . . .(3 - j + 18j - 6j^2)a + (10 + 5j - 4j - 2j^2)a \:=\:2 + 12j + 60 - 24j

. . . . . . . . . . . . . . . . . . . . . (9 + 17j)a + (12 + j)a \:=\:62 - 12j

. . . . . . . . . . . . . . . . . . . . . . . . . . . . .(21 + 18j)a \:=\:62 - 12j

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . a \:=\:\frac{62 - 12j}{21 + 18j}

\text{Rationalize: }\:a \;=\;\frac{62-12j}{21+18j}\cdot\frac{21-18j}{21-18j} \;=\;\frac{1302 - 1116j - 252j + 216j^2}{441 - 324j^2}

. . \text{Therefore: }\:a \;=\;\frac{1086 - 1368j}{765} \quad\Rightarrow\quad \boxed{a\;=\;\frac{362 - 456j}{255}}


Now solve for b.

andrewburton
07-08-2011, 02:28 AM
Guys,
Sorry if i didnt explain to clearly, but thank you soroban as you have got exactly the same answer as I got and then for b, I got the answer (140-j290)/255.

The trouble comes when i try to check that answer by trying to solve the equation using the other terms e.g multiplying through by (5-j2) for equation 2, and (1+j6) into equation 1.

It shouldnt matter which ones I use, but it's giving me different answers each time.

I must be makeing algebraic errors


Thanks guys

andrewburton
07-08-2011, 04:47 AM
Thank you Soroban, I must have been making simple errors. I've worked out the answer now.

I got b as 0.55-j1.14

Cheers guys

mmm4444bot
07-08-2011, 09:59 AM
Okay, I did some research and discovered that electrical engineers use symbol j to represent the imaginary unit because symbol i is already used for electric current; hence, symbol j is a standard that exists outside of mainstream algebra-curricula.

Also, were one to type j*2 or 2j, that would disambiguate j2 from j^2. I, too, have seen many people text j2 to mean j^2.

Thank you everyone, for the clarifications.