Question

solve for real numbers x and y:
(2y-3x) + (3x+4)i =1+ i

Answer 1

\[\Large\bf\sf (2y-3x) + (3x+4)\mathcal i =1+ \mathcal i\]For this equation to hold true, the real part must equal the real part, and the imaginary parts must be equal as well.\[\Large\bf\sf 1\quad=\quad 2y-3x\]\[\Large\bf\sf \mathcal i\quad=\quad (3x+4)\mathcal i\]

Answer 2

Look at the second equation, do you see how to solve for x?

Answer 3

would I distribute the i to be 3xi-4i?

Answer 4

No.
We want to find real values for x.
So we should probably try to find a way to get rid of the i's instead of distributing them.

Answer 5

The i is multiplying those brackets.
Let's do the inverse of that to get rid of the i's.

Answer 6

Start by dividing each side by i.

Answer 7

thats for the 3x+4i=i right?

Answer 8

ya.

Answer 9

would that be 3x+4=i??

Answer 10

No, the i's divide out of each side.\[\Large\bf\sf \frac{\cancel{\mathcal i}}{\cancel{\mathcal i}}\quad=\quad \frac{(3x+4)\cancel{\mathcal i}}{\cancel{\mathcal i}}\]

Answer 11

\[\Large\bf\sf 1\quad=\quad 3x+4\]

Answer 12

oh I see so now I need to solve for both equations separately?

Answer 13

Yes.
This equation will allow you to solve for x.
Then after you have x, you can plug it into the other equation to find your y.

Answer 14

oh so I solved 1=3x+4 and I got -1 so I plug that into 2y-3x=1 to get y?

Answer 15

ya

Answer 16

Thank you so much! :)