Substitution vs Elimination in Solving Systems of Equations
1
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When solving systems of equations, is it more efficient in terms of time to solve it using substitution or elimination, and what are your reasons for saying so?
linear-algebra systems-of-equations
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add a comment |
1
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When solving systems of equations, is it more efficient in terms of time to solve it using substitution or elimination, and what are your reasons for saying so?
linear-algebra systems-of-equations
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add a comment |
1
1
1
$begingroup$
When solving systems of equations, is it more efficient in terms of time to solve it using substitution or elimination, and what are your reasons for saying so?
linear-algebra systems-of-equations
$endgroup$
When solving systems of equations, is it more efficient in terms of time to solve it using substitution or elimination, and what are your reasons for saying so?
linear-algebra systems-of-equations
linear-algebra systems-of-equations
edited Aug 19 '14 at 7:14
asked Aug 19 '14 at 6:15
user170481
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2 Answers
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In my experience, elimination. But usually with systems of equations in three or more variables I would use elimination first then substitution afterwards.
answered Aug 19 '14 at 6:39
MathMajorMathMajor
5,59721436
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Why would you use both methods? Is that to ensure accuracy?
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– user170481
Aug 19 '14 at 7:14
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Not really. For example, suppose we have a system of three equations in three unknowns. It might be easy to use elimination to obtain solutions for one or two of the variables, and then you can "substitute" your results back into another equation to solve for the remaining variable. See "Gauss Jordan Elimination" for a specific algorithm in solving systems of linear equations. I notice that you have tagged the question as "Linear-Algebra". In that case, you are best off by forming an augmented matrix $left( A mid vec b right)$ and bringing it into row reduced echelon form to find $vec x$.
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– MathMajor
Aug 19 '14 at 7:21
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Substitution is faster when you dealing with homogeneous equations and in rest of the cases, elimination is quite faster and efficient.
answered Aug 19 '14 at 6:48
MadhusudanaMadhusudana
86111
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$begingroup$
Sorry, I am a beginner. How are homogenous systems of equations different to normal stystems of equations?
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– user170481
Aug 19 '14 at 7:12
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2 Answers
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2 Answers
2
active
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0
$begingroup$
In my experience, elimination. But usually with systems of equations in three or more variables I would use elimination first then substitution afterwards.
answered Aug 19 '14 at 6:39
MathMajorMathMajor
5,59721436
$endgroup$
$begingroup$
Why would you use both methods? Is that to ensure accuracy?
$endgroup$
– user170481
Aug 19 '14 at 7:14
$begingroup$
Not really. For example, suppose we have a system of three equations in three unknowns. It might be easy to use elimination to obtain solutions for one or two of the variables, and then you can "substitute" your results back into another equation to solve for the remaining variable. See "Gauss Jordan Elimination" for a specific algorithm in solving systems of linear equations. I notice that you have tagged the question as "Linear-Algebra". In that case, you are best off by forming an augmented matrix $left( A mid vec b right)$ and bringing it into row reduced echelon form to find $vec x$.
$endgroup$
– MathMajor
Aug 19 '14 at 7:21
add a comment |
0
$begingroup$
In my experience, elimination. But usually with systems of equations in three or more variables I would use elimination first then substitution afterwards.
answered Aug 19 '14 at 6:39
MathMajorMathMajor
5,59721436
$endgroup$
$begingroup$
Why would you use both methods? Is that to ensure accuracy?
$endgroup$
– user170481
Aug 19 '14 at 7:14
$begingroup$
Not really. For example, suppose we have a system of three equations in three unknowns. It might be easy to use elimination to obtain solutions for one or two of the variables, and then you can "substitute" your results back into another equation to solve for the remaining variable. See "Gauss Jordan Elimination" for a specific algorithm in solving systems of linear equations. I notice that you have tagged the question as "Linear-Algebra". In that case, you are best off by forming an augmented matrix $left( A mid vec b right)$ and bringing it into row reduced echelon form to find $vec x$.
$endgroup$
– MathMajor
Aug 19 '14 at 7:21
add a comment |
0
0
0
$begingroup$
In my experience, elimination. But usually with systems of equations in three or more variables I would use elimination first then substitution afterwards.
answered Aug 19 '14 at 6:39
MathMajorMathMajor
5,59721436
$endgroup$
In my experience, elimination. But usually with systems of equations in three or more variables I would use elimination first then substitution afterwards.
answered Aug 19 '14 at 6:39
MathMajorMathMajor
5,59721436
answered Aug 19 '14 at 6:39
MathMajorMathMajor
5,59721436
answered Aug 19 '14 at 6:39
MathMajorMathMajor
5,59721436
answered Aug 19 '14 at 6:39
MathMajorMathMajor
5,59721436
5,59721436
$begingroup$
Why would you use both methods? Is that to ensure accuracy?
$endgroup$
– user170481
Aug 19 '14 at 7:14
$begingroup$
Not really. For example, suppose we have a system of three equations in three unknowns. It might be easy to use elimination to obtain solutions for one or two of the variables, and then you can "substitute" your results back into another equation to solve for the remaining variable. See "Gauss Jordan Elimination" for a specific algorithm in solving systems of linear equations. I notice that you have tagged the question as "Linear-Algebra". In that case, you are best off by forming an augmented matrix $left( A mid vec b right)$ and bringing it into row reduced echelon form to find $vec x$.
$endgroup$
– MathMajor
Aug 19 '14 at 7:21
add a comment |
$begingroup$
Why would you use both methods? Is that to ensure accuracy?
$endgroup$
– user170481
Aug 19 '14 at 7:14
$begingroup$
Not really. For example, suppose we have a system of three equations in three unknowns. It might be easy to use elimination to obtain solutions for one or two of the variables, and then you can "substitute" your results back into another equation to solve for the remaining variable. See "Gauss Jordan Elimination" for a specific algorithm in solving systems of linear equations. I notice that you have tagged the question as "Linear-Algebra". In that case, you are best off by forming an augmented matrix $left( A mid vec b right)$ and bringing it into row reduced echelon form to find $vec x$.
$endgroup$
– MathMajor
Aug 19 '14 at 7:21
$begingroup$
Why would you use both methods? Is that to ensure accuracy?
$endgroup$
– user170481
Aug 19 '14 at 7:14
$begingroup$
Why would you use both methods? Is that to ensure accuracy?
$endgroup$
– user170481
Aug 19 '14 at 7:14
$begingroup$
Not really. For example, suppose we have a system of three equations in three unknowns. It might be easy to use elimination to obtain solutions for one or two of the variables, and then you can "substitute" your results back into another equation to solve for the remaining variable. See "Gauss Jordan Elimination" for a specific algorithm in solving systems of linear equations. I notice that you have tagged the question as "Linear-Algebra". In that case, you are best off by forming an augmented matrix $left( A mid vec b right)$ and bringing it into row reduced echelon form to find $vec x$.
$endgroup$
– MathMajor
Aug 19 '14 at 7:21
$begingroup$
Not really. For example, suppose we have a system of three equations in three unknowns. It might be easy to use elimination to obtain solutions for one or two of the variables, and then you can "substitute" your results back into another equation to solve for the remaining variable. See "Gauss Jordan Elimination" for a specific algorithm in solving systems of linear equations. I notice that you have tagged the question as "Linear-Algebra". In that case, you are best off by forming an augmented matrix $left( A mid vec b right)$ and bringing it into row reduced echelon form to find $vec x$.
$endgroup$
– MathMajor
Aug 19 '14 at 7:21
add a comment |
0
$begingroup$
Substitution is faster when you dealing with homogeneous equations and in rest of the cases, elimination is quite faster and efficient.
answered Aug 19 '14 at 6:48
MadhusudanaMadhusudana
86111
$endgroup$
$begingroup$
Sorry, I am a beginner. How are homogenous systems of equations different to normal stystems of equations?
$endgroup$
– user170481
Aug 19 '14 at 7:12
add a comment |
0
$begingroup$
Substitution is faster when you dealing with homogeneous equations and in rest of the cases, elimination is quite faster and efficient.
answered Aug 19 '14 at 6:48
MadhusudanaMadhusudana
86111
$endgroup$
$begingroup$
Sorry, I am a beginner. How are homogenous systems of equations different to normal stystems of equations?
$endgroup$
– user170481
Aug 19 '14 at 7:12
add a comment |
0
0
0
$begingroup$
Substitution is faster when you dealing with homogeneous equations and in rest of the cases, elimination is quite faster and efficient.
answered Aug 19 '14 at 6:48
MadhusudanaMadhusudana
86111
$endgroup$
Substitution is faster when you dealing with homogeneous equations and in rest of the cases, elimination is quite faster and efficient.
answered Aug 19 '14 at 6:48
MadhusudanaMadhusudana
86111
answered Aug 19 '14 at 6:48
MadhusudanaMadhusudana
86111
answered Aug 19 '14 at 6:48
MadhusudanaMadhusudana
86111
answered Aug 19 '14 at 6:48
MadhusudanaMadhusudana
86111
86111
$begingroup$
Sorry, I am a beginner. How are homogenous systems of equations different to normal stystems of equations?
$endgroup$
– user170481
Aug 19 '14 at 7:12
add a comment |
$begingroup$
Sorry, I am a beginner. How are homogenous systems of equations different to normal stystems of equations?
$endgroup$
– user170481
Aug 19 '14 at 7:12
$begingroup$
Sorry, I am a beginner. How are homogenous systems of equations different to normal stystems of equations?
$endgroup$
– user170481
Aug 19 '14 at 7:12
$begingroup$
Sorry, I am a beginner. How are homogenous systems of equations different to normal stystems of equations?
$endgroup$
– user170481
Aug 19 '14 at 7:12
add a comment |
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