Solving system of equations with elimination

[Andrew Rubin]

I am doing a course in matrix methods, and I am currently working on solving system of equations by elimination. Pedagogically, I have attended better courses. Due to lack of explanation of concrete steps, I am stuck and hopefully some of you may help me.

We are to solve a system of equations with a vector [MATH]\textbf{x}[/MATH] of unknowns:

1. [MATH]1x_{1} + 10x_{2} + 9x_{3} = 23[/MATH]2. [MATH]7x_{1} + 7x_{2} + 1x_{3} = 35[/MATH]3. [MATH]7x_{1} + 8x_{2} + 1x_{3} = 37[/MATH]
We are told that we will eliminate the variables one-by-one.

Step 1 is to multiply eq. 1 with -7 and add results to eq. 2 and eq. 3:

1. [MATH](-7)\cdot1x_{1} + (-7)\cdot10\cdot(-7)x_{2} + (-7)\cdot9x_{3} = (-7)\cdot23[/MATH]
1. [MATH]=-7x_{1} + -70x_{2} + -63x_{3} = -161 [/MATH]
We have now eliminated [MATH]x_{1}[/MATH] in eq. 2 and 3:

2. [MATH]-63x_{2} + -63x_{3} = -126[/MATH]3. [MATH]-62x_{2} + -63x_{3} = -124[/MATH]
*Here comes the my two questions*

We are told that step 2 is to "take eq. 2 and add a multiple of it to eq. 3 to eliminate [MATH]x_{2}[/MATH]". Out of nowhere, we are told to add [MATH]-62/63[/MATH] times eq. 2 to eq. 3 to eliminate [MATH]x_{2}[/MATH]. I do not understand why exactly this number is chosen, and my attempt at solving this is not correct:

3. [MATH](-62/63)\cdot(-63)x_{2}+(-62)x_{2} + (-62/63)\cdot(-63)x_{3}+(-63)x_{3} = (-62/63)\cdot(-126)+(-124)[/MATH]3. [MATH] = 0x_{2} + (-1)x_{3} = 0[/MATH]
In the course, this is the result:



After this result, we continue with back-solving the system of equations, starting with [MATH]-62/63x_{3}=0[/MATH]. Since I am unable to understand the second step, I am currently stuck.

To sum up:

Q1: How do we determine that [MATH]-62/63[/MATH] is correct here?
Q2: What am I doing wrong in step 2?

 

[lev888]

3. (−62/63)⋅(−63)+(−62)x₂...

You lost variables of eq. 2. It should be (−62/63)⋅(−63)x₂ + ...
The idea here is to get x₂ with a coefficient of equal abs value and opposite sign as the one in eq 3, so that x₂ is eliminated if we add the equations.

Also, why bother with (−62/63) if we can eliminate x₃ just by subtracting the equations?

 

[Andrew Rubin]

You lost variables of eq. 2. It should be (−62/63)⋅(−63)x₂ + ...
The idea here is to get x₂ with a coefficient of equal abs value and opposite sign as the one in eq 3, so that x₂ is eliminated if we add the equations.

Also, why bother with (−62/63) if we can eliminate x₃ just by subtracting the equations?

Thanks! I edited it so it should be correct now. I do not understand the reasoning behind choosing [MATH](-62/63)[/MATH], nor am I sure about the concrete steps in subtraction of the equations. Would you mind describing this? I have seen multiple online step-by-step calculators that arrive at correct results as the course example, but I would like to understand the exact thinking of the instructor as I am only halfway through the course.

 

[Steven G]

This is a problem that I have with how some teachers teach algebra. It is excellent for a student to know how to solve a linear equation but it is important for a student to know how to come up with a linear equation when they need one. This last part is usually not taught.


So you have -63x₂ and want it to become 62x₂.

That is you want to know what to multiply -63 by to get 62. This leads to the equation -63x = 62 and the answer is x=-62/63.

Now I would not solve your problem that way. If I was insistent on eliminating x₂, instead of introducing fractions I would multiply the top equation by 62 and the bottom equation by 63.

Of course the easiest way to proceed at this point is to eliminate x₃

 

[lev888]

Thanks! I edited it so it should be correct now. I do not understand the reasoning behind choosing [MATH](-62/63)[/MATH], nor am I sure about the concrete steps in subtraction of the equations. Would you mind describing this? I have seen multiple online step-by-step calculators that arrive at correct results as the course example, but I would like to understand the exact thinking of the instructor as I am only halfway through the course.

Please look up a tutorial on youtube, etc. E.g. Khan Academy.

 

[Andrew Rubin]

This is a problem that I have with how teachers teach algebra. It is excellent for a student to know how to solve a linear equation but it is important for a student to know how to come up with a linear equation when they need one. This last part is usually not taught.


So you have -63x₂ and want it to become 62x₂.

That is you want to know what to multiply -63 by to get 62. This leads to the equation -63x = 62 and the answer is x=-62/63.

Now I would not solve your problem that way. If I was insistent on eliminating x₂, instead of introducing fractions I would multiply the top equation by 62 and the bottom equation by 63.

Of course the easiest way to proceed at this point is to eliminate x₃

Wow, thanks for the good explanation! Seemed to be a leap there that was not mentioned (maybe the instructor assumes a higher understanding than I am currently at).

Now I appreciate why we use [MATH]-62/63[/MATH] and the idea of getting coefficients of equal value with opposite sign. To really proceed, it would be very helpful with input on how eq. 3 should be changed to arrive at the same result as the course instructor.

 

[Steven G]

1st of all you do NOT have to get the coefficients of equal value with opposite sign. They can be the same sign! You then just subtract instead of adding!

I am not sure what you are asking about eq 3 (you also have more than one equation named 3 !!)

 

[Andrew Rubin]

1st of all you do NOT have to get the coefficients of equal value with opposite sign. They can be the same sign! You then just subtract instead of adding!

I am not sure what you are asking about eq 3 (you also have more than one equation named 3 !!)

Ok, thanks for the clarification. My question concerns getting step 2 correct:

After eliminating [MATH]x_{1}[/MATH] in eq. 2 and eq. 3, we proceed to eliminate [MATH]x_{2}[/MATH] in eq. 3. The proposed solution is to take eq. 2 and add a multiple of it to eq. 3, and the proposed solution is to use [MATH](-62/63)[/MATH]. However, when I follow this procedure I get:

3. [MATH] (−62/63)⋅(−63)x_{2}+(−62)x_{2}+(−62/63)⋅(−63)x_{3}+(−63)x_{3}=(−62/63)⋅(−126)+(−124) \implies 0x_{2}+(−1)x_{3}=0[/MATH]
I think the rest of the solving process makes sense both before and after this point, but I am currently not finding a good solution to this.

 

[pka]

For the life of me I cannot see why some still worship busywork.
LOOK HERE. At least final answers can be checked.