Solve for x&y , complex numbers

J

Joni

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Joined
Dec 31, 2009
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6
I'm new so...here is my equation,
Need to solve for x and y (3x-1)+(y+5)i=1-3i

Here is what I have and I know it is not correct

3x+y-6i=1-3i
3x-6i+1-3i=y this is not correct according to my book so I am lost cannot see the steps to x and y. Please can you show me the complete steps.
 
\(\displaystyle Assuming\ x\ and\ y\ are\ real.....\)

\(\displaystyle (3x-1)=1\ and\ y+5=-3.\)
 
Criss is it possible to show me the steps in how you arrived to that answer
 
\(\displaystyle Hi,\ Joni,\)
\(\displaystyle Complex\ numbers\ are\ strange\ in\ the\ beginning.\)
\(\displaystyle They\ have\ 2\ components,\ a\ real\ part\ and\ imaginary\ part.\)
\(\displaystyle The\ real\ part\ is\ the\ sum\ of\ the\ real\ numbers.\)
\(\displaystyle The\ imaginary\ part\ is\ the\ sum\ of\ the\ i\ parts.\)

\(\displaystyle They\ are\ already\ grouped\ properly\ in\ your\ example.\)

\(\displaystyle You\ have\ y+5\ of\ the\ imaginary\ part\ which\ must\ be\ -3.\)
\(\displaystyle Your\ equation\ is\ already\ written\ as\......real\ part\ +\ imaginary\ part.\)
\(\displaystyle That\ means\ the\ 1\ on\ the\ right\ is\ 3x-1\ and\ the\ -3\ on\ the\ right\ is\ y+5.\)
 
OK I understand what you are saying although when I solve for x in the 3x-1 I get 1/3 not 1
 
\(\displaystyle You\ need\ to\ try\ that\ again, Joni.\)
\(\displaystyle 3x-1=1\)

\(\displaystyle (3x-1)+1=(1)+1\ this\ gets\ 3x\ by\ itself.\)

\(\displaystyle 3x=2\)

\(\displaystyle \frac{3x}{3}=\frac{2}{3}\ gets\ the\ value\ of\ x.\)

\(\displaystyle Or\ 3x-1=1\ means\ 3x\ is\ 2,\ since\ 2-1=1.\)
 
I just figured it out Chris and I completly understnd now how this one plays out.
 
Re: Linear Combination - Multiplication Method

this is what i have so far. is this right, so far? please show me all the steps

3x - 5y = 13
+ x - 2y = 5 x 3

3x - 5y = 13
- 3x - 6y = 15

-11 y = -2 divide both sides by -11

y = 2/11

3x - 5y = 13
+ 5y + 5y
 
What happened on line 4 ?

You went from an x to a -x but held the y sign the same.

Is your typo on line 2 or 4 ?

I think what you meant was subtract the entire line 4 from line 3.

You are nearly there...
You should have had +6y on that line as -(-) is +.

That gives 6y-5y=y=-2.

Then use this y to find x from 3x-5(-2)=13,
instead of starting all over again to find x.
 
Re: Linear Combinations - Multiplication Method

is anyone online to answer my previous question. i need it answered now
 
these are the changes you where referring to. aren't they?

3x - 5y = 13
- 3x + 6y = 15
6y - 5y = -2
y = -2

3x - 5 (-2) = 13

3x + 10 = 13
- 10 - 10

3x = 3 div. both sides by 3

x = 1

(1, -2)

is this correct?
 
Re: Linear Combinations - Multiplication Method

i changed the x to a negative 3x because either the x or the y needs to cancel
 
Re: Linear Combinations - Multiplication Method

this is another problem:

7x + 2y = -1 x2
+ 3x - 3y = 19

14x + 4y = -1
+ 3x - 3y = 19

i got stuck here
 
Yes,

now it's possible for you to be sure if you are correct or not.

There is only one pair of (x,y) that can be placed in both original equations.

This corresponds to the point of intersection of two lines with different slopes.

There is only one such (x,y) point.

If you put x=1, y=-2 into the first equation, is it true ? is it 13?

If you put them into the 2nd one, is it still true ?

You see that they are,
hence your solution is correct.

Sorry for the delay, I was out in the snow...
 
Re: Linear Combinations - Multiplication Method

homeworkhelp said:
7x + 2y = -1 x2
+ 3x - 3y = 19

14x + 4y = -1
+ 3x - 3y = 19

i got stuck here
Click to expand...

You're stuck because you haven't gotten the hang of what's going on yet....

7x+2y = -1
3x-3y = 19

first line multiplied by 3 and second line multiplied by 2.
This works because we will have 6y and -6y.

21x+6y = -3 ......... don't forget to multiply the other side.
Remember 7x+2y is -1
3 times (7x+2y) is 3 times (-1)

21x+6y = -3
6x-6y = 38 ...... everything multiplied by 2.

add

27x = 35

\(\displaystyle x = \frac{35}{27}\)

You could now use this x to find y

\(\displaystyle (6)\frac{35}{27}-6y = 38\)

\(\displaystyle (2)\frac{35}{9}-6y = 38\)

\(\displaystyle (2)\frac{35}{9} = 38+6y\)

\(\displaystyle \frac{70}{9}-38 = 6y\)

\(\displaystyle y=\frac{\frac{70}{9}-38}{6}\)

Or do the process of cancelling again instead.
 
Re: Linear Combinations - Multiplication Method

homeworkhelp said:
i changed the x to a negative 3x because either the x or the y needs to cancel
Click to expand...

No!!!

You can't just change the sign of a value within a group of terms without checking for violation.
You have to change the sign of all terms....

x+2y=-4
-x-2y=4.

3x-y=2
-3x+y=-2
 
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