Solve for x and y: Calculating Equations with Unknown Variables

  • Thread starter Thread starter rabihtawil
  • Start date Start date
  • Tags Tags
    Calculation
AI Thread Summary
The discussion revolves around solving a system of linear equations with unknown variables. The user initially struggles with understanding how to derive the solutions for the equations 2x + y + 3 = 0 and x + 2y - 3 = 0. After some confusion, they realize that the problem can be approached using an augmented matrix format. The correct representation of the system is provided, leading to the conclusion that solving these equations is essential for mastering the topic. The user expresses a desire to improve their math skills, particularly as a computer science student facing challenges in their remaining courses.
rabihtawil
Messages
10
Reaction score
0

Homework Statement



i have the answer for this i just don't know how it happened
Issue1:
2x+y+3=0
x= -3

Issue2:
x+2y-3=0
y=3

Homework Equations



How did they come up with this solution?

The Attempt at a Solution



I reread it & try to solve it no luck :(
 
Physics news on Phys.org
oh never mind i figured it out, it's a system of linear equation

2 1 3
1 2 -3

solve the system the above results will come up. thank you forum
 
You know how to solve matrices but not a system of 2 linear equations in 2 variables?
You have essentially started with the harder stuff and skipped the easier ones :bugeye:
 
My math skills are bad, I'm just working hard to pass these classes because I'm a computer science student. i finished all my major except 3 math courses which i am having the worst experience of my life in.
 
rabihtawil said:
oh never mind i figured it out, it's a system of linear equation

2 1 3
1 2 -3

solve the system the above results will come up. thank you forum
If you're working with an augment matix, it should be
Code:
[2 1 | -3]
[1 2 | 3]

The rows in this matrix come from these equations:
2x + y = -3
x + 2y = 3
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top