Find x and y if x, y are members of real numbers and: (x+i)(3-iy)=1+13i
I first expanded it to give: 3x-yix+3i+y=1+13i
Then I equaled 3x+y=1 and -yx+3=13
But afterwards I do other steps and get the wrong answer.
Find x and y if x, y are members of real numbers and: (x+i)(3-iy)=1+13i
I first expanded it to give: 3x-yix+3i+y=1+13i
Then I equaled 3x+y=1 and -yx+3=13
But afterwards I do other steps and get the wrong answer.
Last edited by MarkFL; January 9th, 2017 at 22:10. Reason: Include Problem In Post
Your approach and posted work are ok.
To continue,
$$3x+y=1\implies y=1-3x$$
$$3-xy=13\implies xy=-10$$
$$x(1-3x)=-10$$
$$(3x+5)(x-2)=0$$
$$(x,y)=\left(-\frac53,6\right),(2,-5)$$
Does that match your results?