How do you solve for x using the expansion method?

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Homework Help Overview

The original poster presents a system of equations involving three variables, x, y, and z, and seeks to solve for x using a method referred to as the "expansion method." The equations provided are x + y + z = 1, x + y + 2z = 2, and x + y + 3z = 1.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the original poster's approach to calculating the determinant of a matrix derived from the system of equations. Some express confusion about the "expansion method" and its application, while others suggest alternative methods such as simple subtraction to analyze the system.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and the method mentioned. Some have provided insights into the determinant calculation, noting that it results in zero, which implies no unique solution exists for x. However, there is no consensus on the terminology or the method itself.

Contextual Notes

There is uncertainty regarding the definition and application of the "expansion method," and participants are questioning the assumptions made in the original poster's approach. Additionally, the presence of multiple potential values for z raises questions about the system's solvability.

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Homework Statement




solve for x using the expansion method

x + y + z =1
x + y + 2z =2
x + y + 3z = 1

Homework Equations


none


The Attempt at a Solution



1 1 1 1 1
1 1 2 1 1
1 1 3 1 1


1 1 1 1 1
1 1 2 1 1
1 1 3 1 1


1 1 1 1 1
1 1 2 1 1
1 1 3 1 1

(1)(1)(3) + (1)(2)(1) + (1)(1)(1) - (3)(1)(1) + (1)(2)(1) + (1)(1)(1)
(3)+(2)+(1) - (3)(2)(1)
=6 - 6
= 0

would this be undefined then?
 
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Hi haengbon! :smile:

I'm not familiar with the "expansion method", but simple subtraction on …
haengbon said:
x + y + z =1
x + y + 2z =2
x + y + 3z = 1

… gives both z = 1 and z = 0, so clearly there are no solutions! :wink:
 
tiny-tim's not alone. I haven't heard of this method, either, so the work you show is a complete mystery to me.
 
I think the OP is writing the system in terms of a matrix A where Ax=b and is trying to calculate the determinant of A. In this case, one gets

det A = (1)(1)(3)+(1)(2)(1)+(1)(1)(1)-(1)(1)(1)-(1)(2)(1)-(1)(1)(3) = 3+2+1-1-2-3 = 0

so there's no unique solution for x.

Perhaps "expansion method" refers to Cramer's rule.
 
That's the method that I learned for calculating the determinate in HS. You copy the first column over on the Right side so the diagonals are all nice and in line...then you calculate the number as shown above...forgot the + signs on the last line there.
(3)+(2)+(1) - (3)(2)(1) should have + in between the last 3 2 1.
 

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