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Solving Linear Systems (Basic Question)

by Atran
Tags: basic, linear, solving, systems
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Atran
#1
Jan3-14, 05:41 PM
P: 84
Hi. I don't understand how a solution to a linear system is obtained (for example geometrically; don't consider the substitution method and elimination), and I am feeling very frustrated.

Say I have the following equations:
y = x + 5
b = 2*a (the relation remains the same even if I change the variables)
Obviously, the solution is (x=a=5 and y=b=10) or (x=b=-5 and y=a=-10).

The first equation has the solution sets, A1={(x, x+5) : x∈R} and A2={(x+5, x) : x∈R}.
The second equation has the solution sets, B1={(x, 2*x) : x∈R} and B2={(2*x, x) : x∈R}

A1 ∩ B1 and A2 ∩ B2 are the solution sets to the system, if x=a and y=b.
A1 ∩ B2 and A2 ∩ B1 are the solution sets to the system, if x=b and y=a.

How can I prove that the variables which are meant to be equal (for example x=a), must be both the first or the second element in given pairs? For example if x=a and I consider A2, then I must consider B2, i.e. if x is the second element of the pairs x and y, then a must also be the second element of pairs a and b.

Am I thinking right? I've been looking at many websites but none really cleared up my confusion.
I'm really thankful if you can explain this clearly.
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Mark44
#2
Jan3-14, 06:14 PM
Mentor
P: 21,258
Quote Quote by Atran View Post
Hi. I don't understand how a solution to a linear system is obtained (for example geometrically; don't consider the substitution method and elimination), and I am feeling very frustrated.

Say I have the following equations:
y = x + 5
b = 2*a (the relation remains the same even if I change the variables)
Obviously, the solution is (x=a=5 and y=b=10) or (x=b=-5 and y=a=-10).
The two equations involve different pairs of variables, so the two equations aren't related at all. If the second equation happened to be y = 2x, then you would have two lines that intersect. The intersection point would be the solution of the system of equations.
Quote Quote by Atran View Post

The first equation has the solution sets, A1={(x, x+5) : x∈R} and A2={(x+5, x) : x∈R}.
The second equation has the solution sets, B1={(x, 2*x) : x∈R} and B2={(2*x, x) : x∈R}
By convention we write the ordered pairs as (x, y), in that order. The solution set would be {(x, y) | y = x + 5, x ##\in## R}, and similar for the second equation.
Quote Quote by Atran View Post

A1 ∩ B1 and A2 ∩ B2 are the solution sets to the system, if x=a and y=b.
A1 ∩ B2 and A2 ∩ B1 are the solution sets to the system, if x=b and y=a.
Why are you doing this? By using a different set of variables (a and b), you are overcomplicating what is a simple problem.
Quote Quote by Atran View Post

How can I prove that the variables which are meant to be equal (for example x=a), must be both the first or the second element in given pairs? For example if x=a and I consider A2, then I must consider B2, i.e. if x is the second element of the pairs x and y, then a must also be the second element of pairs a and b.
You don't need to prove this, as the variables appear in a certain order by convention. If you write the system as
y = x + 5
y = 2x

then the system is trivial to solve.
Quote Quote by Atran View Post

Am I thinking right? I've been looking at many websites but none really cleared up my confusion.
I'm really thankful if you can explain this clearly.
1MileCrash
#3
Jan3-14, 06:15 PM
1MileCrash's Avatar
P: 1,295
I don't understand how those two equations form a linear system, or your "solution."


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