Solve The System Of Linears Equations for x and y

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In summary, the equations (\cos \theta )x + (\sin \theta )y = 1 and (-\sin \theta )x + (\cos \theta )y = 0 can be solved using parametric equations, with x = \cos \theta and y = \sin \theta as the final solution. The equations can also be solved by multiplying the equations by appropriate values and adding them together to get x = \cos \theta and y = \sin \theta.
  • #1
Bashyboy
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Homework Statement


[itex](\cos \theta )x + (\sin \theta )y = 1[/itex]

and

[itex](-\sin \theta )x + (\cos \theta )y = 0[/itex]

Homework Equations





The Attempt at a Solution



Evidently the answer is that [itex]x = \cos \theta[/itex] and that [itex]y = \sin \theta[/itex].

Here is my work:

[itex]x = \frac{1 - (\sin \theta )y}{\cos \theta}[/itex]

Substituting this into the second equation, and simplifying:

[itex]y = \frac{\tan \theta}{\sin \theta tan \theta + \cos \theta}[/itex]

I then took this equation and back-substituted into [itex]x = \frac{1 - (\sin \theta )y}{\cos \theta}[/itex], hoping that everything would simplify such that [itex]x= \cos \theta[/itex]; however, things began to look quite messy. How am I to solve this problem?
 
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  • #2
Bashyboy said:

Homework Statement


[itex](\cos \theta )x + (\sin \theta )y = 1[/itex]

and

[itex](-\sin \theta )x + (\cos \theta )y = 0[/itex]

Homework Equations





The Attempt at a Solution



Evidently the answer is that [itex]x = \cos \theta[/itex] and that [itex]y = \sin \theta[/itex].

Here is my work:

[itex]x = \frac{1 - (\sin \theta )y}{\cos \theta}[/itex]

Substituting this into the second equation, and simplifying:

[itex]y = \frac{\tan \theta}{\sin \theta tan \theta + \cos \theta}[/itex]

Multiply the numerator and denominator of that last equation by ##\cos\theta## and you will have it. Much easier to use determinants in the first place though.
 
  • #3
Personally, I would not have done the problem that way. Starting from the original equations,
[itex]cos(\theta)x+ sin(\theta)y= 1[/itex] and
[itex]-sin(\theta)x+ cos(\theta)y= 0[/itex]

Multiply the first equation by [itex]cos(theta)[/itex] and the second equation by [itex]-sin(\theta)[/itex] to get
[itex]cos^2(\theta)x+ sin(\theta)cos(\theta)y= cos(\theta)[/itex]
[itex]sin^2(\theta)x- sin(\theta)cos(\theta)y= 0[/itex]
and then add:
[itex]x= cos(\theta)[/itex]

Then multiply the first equation by [itex]sin(\theta)[/itex] and the second equation by [itex]cos(\theta)[/itex]to get [itex]sin(\theta)cos(\theta)x+ sin^2(\theta)y= sin(\theta)[/itex]
[itex]-sin(\theta)cos(\theta)x+ cos^2(\theta)y= 0[/itex]
Adding gives [itex]y= sin(\theta)[/itex].
 
  • #4
And so theta will be the parameter to the parametric equations that represent the solution?
 

FAQ: Solve The System Of Linears Equations for x and y

How do I solve a system of linear equations for x and y?

To solve a system of linear equations for x and y, you can use the substitution method or the elimination method. In the substitution method, you solve one equation for either x or y and then substitute that value into the other equation. In the elimination method, you manipulate the equations to eliminate one variable and then solve for the remaining variable.

What is the purpose of solving a system of linear equations?

The purpose of solving a system of linear equations is to find the values of x and y that satisfy both equations simultaneously. These values represent the coordinates of the point where the two lines intersect and can be used to solve real-world problems involving two variables.

Can a system of linear equations have more than one solution for x and y?

Yes, a system of linear equations can have one, infinite, or no solutions. If the two equations represent two parallel lines, there is no point of intersection and the system has no solution. If the two equations represent the same line, there are infinite solutions as every point on the line satisfies both equations. If the two equations represent intersecting lines, there is one unique solution.

What should I do if I get a contradiction while solving a system of linear equations?

If you get a contradiction, such as 0 = 5, while solving a system of linear equations, it means that the system has no solution. This can happen if the two equations represent parallel lines, or if there is an error in your calculations.

Can I solve a system of linear equations with more than two variables?

No, a system of linear equations can only be solved if there are the same number of equations as variables. If there are more than two variables, additional equations are needed to solve for each variable. These types of systems are known as simultaneous equations and require advanced techniques to solve.

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