Solving Linear Systems (Basic Question)

In summary, the conversation discusses the frustration with understanding how to obtain a solution to a linear system, specifically through geometric means without using the substitution method or elimination. The equations y = x + 5 and b = 2*a are given as an example, with the solution being (x=a=5 and y=b=10) or (x=b=-5 and y=a=-10). The conversation also delves into the solution sets for each equation and the intersection of these sets as the solution to the system. The individual also questions the convention of writing ordered pairs and seeks clarification on their understanding.
  • #1
Atran
93
1
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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  • #2
Atran said:
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.
Atran said:
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.
Atran said:
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.
Atran said:
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.
Atran said:
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.
 
  • #3
I don't understand how those two equations form a linear system, or your "solution."
 

Related to Solving Linear Systems (Basic Question)

1. What is a linear system?

A linear system is a set of two or more linear equations that are solved simultaneously to find the values of the variables that satisfy all of the equations.

2. How do you solve a linear system?

There are several methods for solving a linear system, including substitution, elimination, and graphing. The most commonly used method is elimination, where you eliminate one variable by adding or subtracting the equations from each other until only one variable remains.

3. Can a linear system have more than one solution?

Yes, a linear system can have one, infinite, or no solutions. One solution exists when the two equations intersect at a single point, infinite solutions exist when the two equations are equivalent and overlap, and no solutions exist when the two equations are parallel and do not intersect.

4. Why is solving linear systems important?

Solving linear systems is important because it allows us to find the values of variables in real-world situations, such as in business, engineering, and science. It also helps us to understand the relationships between different variables and make predictions based on those relationships.

5. Are there any tips for solving linear systems more efficiently?

One helpful tip is to always check your solution by substituting the values into each equation to ensure that they satisfy all of the equations. Additionally, it can be helpful to solve for one variable first and then substitute that value into the other equation to solve for the remaining variable.

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