Destinyyyy
  • Destinyyyy
perform the addition.. How do I solve this? The answer I got was incorrect.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Destinyyyy
  • Destinyyyy
(-5+5i) + (9+8i)
anonymous
  • anonymous
4+13i
Destinyyyy
  • Destinyyyy
Yes I know that thats the correct answer. Im not asking for that. Im asking how to solve it

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More answers

DecentNabeel
  • DecentNabeel
\[\left(5i-5\right)+\left(8i+9\right)\] \[\mathrm{Remove\:parentheses}:\quad \left(a\right)=a\] \[=\left(5i-5\right)+8i+9\] \[\mathrm{Add/Subtract\:the\:numbers:}\:-5+9=4\] \[=8i+5i+4\] \[\mathrm{Add\:similar\:elements:}\:8i+5i=13i\] \[=13i+4\] that is the answer
DecentNabeel
  • DecentNabeel
are you understand @Destinyyyy
Destinyyyy
  • Destinyyyy
Sorry one second while I read it.. There was a spider in my room :(
DecentNabeel
  • DecentNabeel
hahaha no problem @Destinyyyy
Destinyyyy
  • Destinyyyy
My example said to remove the i and put it outside of the equation.. (-5 +5) +(9+8)i
Destinyyyy
  • Destinyyyy
It was fluttering huge... Death by broom
Destinyyyy
  • Destinyyyy
it change f-ing to fluttering -.-
DecentNabeel
  • DecentNabeel
\[\mathrm{Simplify}\:\left(-5+5\right):\quad 0\] \[=i\left(8+9\right)+0\] \[\mathrm{Simplify}\:i\left(8+9\right):\quad 17i\] 17i+0=17i
anonymous
  • anonymous
The way your book does it is still true but i think its in this format (a+bi) + (d+ ci) = (a+d) (b+c) i
anonymous
  • anonymous
To add or subtract we use Add (a+bi) + (c +di) = (a+b) + (c +d)i Subtract (a+bi) (c +di) = (a+b) - (c +d)i So in your case (-5+5i) + (9+8i) = (-5+5) + (9+8)i -5+9 + 5+8 4 + 13i
DecentNabeel
  • DecentNabeel
any confusion @Destinyyyy
DecentNabeel
  • DecentNabeel
@Deeezzzz -5+9=4 not -4
Destinyyyy
  • Destinyyyy
Okay.. I get it now.. I multiplied the parenthesis... Thank you everyone!!
anonymous
  • anonymous
Makes sense?
Destinyyyy
  • Destinyyyy
Yes.. Can I get help with another one?
DecentNabeel
  • DecentNabeel
yes tell me @Destinyyyy
anonymous
  • anonymous
Thanks! @DecentNabeel simple arithmetic mistake
DecentNabeel
  • DecentNabeel
no problem @Deeezzzz
Destinyyyy
  • Destinyyyy
Perform the addition (-9/2 +1/2i) + (3/2 - 7/2i)
DecentNabeel
  • DecentNabeel
\[\left(-\frac{9}{2}+\frac{i}{2}\right)+\left(\frac{3}{2}-\frac{7i}{2}\right)\] \[\mathrm{Remove\:parentheses}:\quad \left(a\right)=a\] \[=\left(-\frac{9}{2}+\frac{i}{2}\right)+\frac{3}{2}-\frac{7i}{2}\] \[\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}\] \[=\frac{-7i+i+3-9}{2}\] refine \[=\frac{-6i-6}{2}\] factor out -6 \[=-\frac{6\left(i+1\right)}{2}\] \[\mathrm{Divide\:the\:numbers:}\:\frac{6}{2}=3\] \[=-\left(3\left(i+1\right)\right)\] \[\mathrm{Negate}\:-\left(3\left(i+1\right)\right)=-3\left(i+1\right)\] \[=-3\left(i+1\right)\] that is the answer
DecentNabeel
  • DecentNabeel
ok @Destinyyyy
Destinyyyy
  • Destinyyyy
??? Why is the 1 now i and the 7 now 7i?
Destinyyyy
  • Destinyyyy
My examples show to put the i on the outside of the equation like @Nixy did
DecentNabeel
  • DecentNabeel
-3i+3
Destinyyyy
  • Destinyyyy
Still confused
DecentNabeel
  • DecentNabeel
why confused tell me
DecentNabeel
  • DecentNabeel
you said i on the out sides .. so that.. 3(i+1) =3i+3
Destinyyyy
  • Destinyyyy
No I mean like this (-9/2 +1/2) + (3/2 - 7/2)i for the first step
DecentNabeel
  • DecentNabeel
\[\mathrm{Remove\:parentheses}:\quad \left(a\right)=a\] \[=\left(-\frac{9}{2}+\frac{i}{2}\right)+\frac{3}{2}-\frac{7i}{2}\]
DecentNabeel
  • DecentNabeel
alright @Destinyyyy
Destinyyyy
  • Destinyyyy
I dont understand why 1 is now i and why its 7i
DecentNabeel
  • DecentNabeel
ok i expla it
anonymous
  • anonymous
\( \huge (-\frac{9}{2} +\frac{1}{2}i) + (\frac{3}{2} - \frac{7}{2}i) \) Is the above how it looks in your book?
DecentNabeel
  • DecentNabeel
(-9/2 +1/2i) + (3/2 - 7/2i) that is your question
anonymous
  • anonymous
\( \huge (-\frac{9}{2} +\frac{1}{2}i) + (\frac{3}{2} - \frac{7}{2}i) \) \( \huge (-\frac{9}{2} +\frac{1}{2}) + (\frac{3}{2} - \frac{7}{2}) i\) \( \huge (-\frac{9}{2} +\frac{3}{2}) + (\frac{1}{2} - \frac{7}{2}) i\) \( \huge -\frac{6}{2} + (-\frac{6}{2}) i\) \( \huge -3 -3i\)

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