MHB Find x and y if x, y are members of real numbers and: (x+i)(3-iy)=1+13i

AI Thread Summary
To solve for x and y in the equation (x+i)(3-iy)=1+13i, the initial expansion leads to two equations: 3x + y = 1 and -xy + 3 = 13. From the first equation, y can be expressed as y = 1 - 3x. Substituting this into the second equation results in a quadratic equation, which factors to (3x + 5)(x - 2) = 0. The solutions yield the pairs (x, y) = (-5/3, 6) and (2, -5), confirming the calculations.
Gladier
Messages
1
Reaction score
0
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 a moderator:
Mathematics news on Phys.org
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?
 

Thread 'Trigonometry problem of interest'

Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
View full post »
Thread 'Video on imaginary numbers and some queries'

Thread 'Video on imaginary numbers and some queries'

Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
View full post »
Thread 'Unit Circle Double Angle Derivations'

Thread 'Unit Circle Double Angle Derivations'

Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
View full post »

Similar threads

MHB What is the Sum of x, y, and z in a Non-Negative Real Number System?
Replies
4
Views
1K
MHB System of Equations: Find Triples $(x,y,z)$
Replies
1
Views
1K
B How can I properly distribute fractions when solving a set of linear equations?
Replies
7
Views
2K
I ##x\lt y \lt z## positive integer numbers and if ##\dfrac 1x+\dfrac 1y+\dfrac 1z=\dfrac 13##
Replies
6
Views
1K
MHB What is the range and inverse of the function y = 1/x?
Replies
3
Views
1K
MHB Find the possible values of k: x+1/y=y+1/z=z+1/x=k
Replies
2
Views
2K
MHB Challenge Problem #7: Σ(x/(y^3+2))≥1
Replies
2
Views
2K
I On base-b expansion of nonnegative reals
Replies
7
Views
2K
MHB Maximize Σ(x^2+y)(y^2+x))/((x+y-1)^2)
Replies
2
Views
1K
MHB Solve for $x$ and $y$: Real Number Condition
Replies
9
Views
2K

Hot Threads

Recent Insights

Back
Top