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Solving a system of three linear equations in 3 unknowns using Gaussian Elimination (or Gauss-Jordan algorithm) in 5 stages. Solutions are all integers. Introductory question where the numbers come out quite nice with not much dividing. Set-up is meant for formative assessment. Adapated from a question copied from Newcastle.
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Scotland schools
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Andrew Brown | said | Ready to use | 2 years, 6 months ago |
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Andrew Brown 2 years, 6 months ago
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Created this as a copy of Gaussian elimination to solve a system of linear equations - Introductory.Name | Status | Author | Last Modified | |
---|---|---|---|---|
Gaussian elimination to solve a system of linear equations - Introductory | Ready to use | Julia Goedecke | 25/08/2021 12:57 | |
Gaussian elimination to solve a system of linear equations - scaffolded | Ready to use | Julia Goedecke | 25/08/2021 12:57 | |
Gaussian elimination to solve a system of equations | Ready to use | Andrew Brown | 12/09/2022 14:48 | |
Gaussian elimination to solve a system of linear equations | Ready to use | Andrew Brown | 12/09/2022 14:48 |
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Name | Type | Generated Value |
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a | integer |
5
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c | integer |
3
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b | integer |
4
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c3 | integer |
1
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c2 | integer |
3
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y | number |
-428
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x | number |
337
|
||||
c1 | integer |
5
|
||||
z | number |
86
|
Generated value: integer
This variable doesn't seem to be used anywhere.
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Transform the equations using logarithms, and introducing the substitutions X=lnx, Y=lny and Z=lnz.
For each non-linear equation in the system, write down the linear equation you determine using logarithms
x{a}y{ab−1}z{a2b−a−ab}=e{c2} | ⟶ | |
x{ac}y{cb}z=e{c1} | ⟶ | |
xy{b}z{ba−b}=e{c3} | ⟶ |
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