Complex Numbers : Equations - Equating real imag. parts : ExamSolutions
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- Опубліковано 7 лют 2012
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I used to take an F to maths, then i found exam solutions.
thank you!!
excellent work mr !!
awesome video, thanks a lot !
good stuff as always.
nice
got both of them right YUS
Sir in the first question can't we also multiply the second one by 4 (4a-4b=-4) and add it to the first (5a-4b=-1) which would give 9a=-5 and find a to be -5/9. And b to be 4/9. Maybe there are 4 end result? Idk might be doing something wrong here that i can't see now.
Minus b add 2b is 1
Cheers
sir, what if we had simultaneous equations ? for example say
2ix - 1.5 y = 2
-4ix + 8.5y = 6
thank you
double the first equation to give 4ix-6y=4 and then add this to the second to give 5.5y=10 so y=20/11. Sub into say the first equation to get x. I leave that to you to try.
sir, if i double the first equation then i get 4ix-3y=4 , right?
Emad Izadi Oops, sorry, you are correct.
sir how would i solve this by using graphs ?, i know that their intersection is the answer but i have the term i so it really makes me confused
you sound like Daniel Ricciardo
Real = real
Imaginary = imaginary
sir, i asked that question from my teacher , and he could not solve it , can you please give me hand for that question? i appreciate that :(
I used to get C at math, but then I took an i to the knee
Igor Aherne Replying to that Skyrim reference after 5 years
@@zxyde After 8 years...