Solving Fractions in Linear Equations: Help Needed!

AI Thread Summary
The discussion revolves around solving a system of linear equations and the confusion regarding obtaining fractions instead of whole numbers. The user initially struggles with the elimination method, specifically when eliminating the variable Z, leading to incorrect fractional results. Other participants suggest showing the work to identify arithmetic errors, as the roots of the equations should yield whole numbers. The user acknowledges a mistake and expresses a desire to master the current material before moving on. Ultimately, the conversation highlights the importance of careful arithmetic and method selection in solving linear equations.
Raizy
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Delete please

Please delete.

I just want to know, if there are anyone out there who knows why I keep getting fractions after adding 2 equations to solve for one of the variables. For example, Z or Y or X would equal 23/20 and the book's answer would be a whole number.

(Edited on Friday 13th 2:53 p.m.)

Homework Statement



ANYTHING that looks similar to solving these by addition/elimination:

Equation One: 3x - y + 2z = 1
Equation Two: 2x +3y +3z = 4
Equation Three: x + y - 4z = -9

The book's answers: (-1, 0, 2)

The attempt at a solution

Okay here is what I did in order to observe: Re-do the problem 3 times in 3 different ways.

Method 1: Eliminate X first -- in which I got the correct answers.
Method 2: Eliminate Y first -- in which I got the correct answers.
Method 3: Booboo. I got a fraction when the first Y was solved for. I got y = 42/116 or 21/58

Comments: So why does it get messed up when I decide to eliminate Z first? How do I know which variable I should eliminate first in order to get the correct answer?

Step 1. Add equation 1 to equation 2:

1a.--> 3(3x - y + 2z) = 3(1)
1b.--> 9x -3y + 6z = 3

2a.--> -2(2x +3y +3z) = -2(4)
2b.--> -4x -6y -6z = -8

Add equation 1b and 2b to get equation 4: 5x -9y = -8 (Edit at 3:52 p.m. Okay... hmm I just picked this error up...

Step 2: Add equation 1 to equation 3:

1a.--> -4(3x -y +2z) = -4(1)
1c.--> -12x +4y -8z = -4

3a.--> -2(x +y -4z) = -2(-9)
3c.--> -2x -2y +8z = 18

Add equation 1c and 2c to get equation 5: -14x +2y = 14

Step 3: Treat equations 4 and 5 as if it were a system of two systems of linear equations.

Add equation 4 to 5:

4a.--> -14(5x -9y) = -14 (-8)
4b.--> -70x +126y = 112

5a.--> -5(-14x +2y) = -5(14)
5b.--> 70x -10y = -70

Add equation 4b to 5b to get the Y variable: Y=42/116 or 21/58COMMENTS:

Does it matter which variable you choose to solve for? What concepts could I be missing? :cry: I've re-read the instructions several times -- I've spent 3 days already trying to re-try with no luck. I am sure, unless my eyes are playing tricks on me, is that it is stated it should not matter which variable you eliminate first as you will always get the same answer.
 
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Hi Raizy,

Can you show how you are trying to solve this? If you show us, we can help you pinpoint where the problem is.
 


Try eliminating the 'y' using the last two equations. You would then have a system of two equations and two unknowns.

Do you want to try matrix operations instead?
 


No one can tell what you are doing wrong if you don't show us what you are doing! The roots of these equations are, in fact, simple whole numbers. It's probably most likely that you are just making arithmetic errors. Perhaps you are not subtracting negative numbers correctly. That's a very common error.
 


yeah, sorry folks for not showing my work. I was assuming I had some universal error which always resulted me getting fractions after solving for the first X Y or Z for systems of linear equations in 3 variables questions. I am currently in class right now so I'll bump this thread next time I stumble on a problem.
 


symbolipoint said:
Try eliminating the 'y' using the last two equations. You would then have a system of two equations and two unknowns.

Do you want to try matrix operations instead?

Impossible, I really don't like reading ahead in the book without mastering the previous section. Unless this is one of those cases where my life would get easier?

FINAL EDIT: Okay, I did a simple mistake... I think I got it (hopefully).
 
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I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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