Solving Fractions in Linear Equations: Help Needed!

In summary: Equation One: 3x - y + 2z = 1Equation Two: 2x +3y +3z = 4Equation Three: x + y - 4z = -9The book's answers: (-1, 0, 2)The attempt at a solutionOkay 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
  • #1
Raizy
107
0
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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  • #2


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.
 
  • #3


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?
 
  • #4


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.
 
  • #5


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.
 
  • #6


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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Related to Solving Fractions in Linear Equations: Help Needed!

1. What are linear equations?

Linear equations are mathematical expressions that involve two variables, typically represented by x and y, and can be written in the form of y = mx + b. They describe a straight line when graphed and are commonly used to model relationships between quantities.

2. How do I solve a linear equation with fractions?

To solve a linear equation with fractions, you can use the following steps:1. Simplify the fractions by finding their common denominators.2. Use the distributive property to eliminate any parentheses.3. Combine like terms on each side of the equation.4. Isolate the variable by adding or subtracting constants on both sides.5. Divide both sides by the coefficient of the variable to solve for the variable.

3. Can I cross-multiply when solving linear equations with fractions?

Yes, you can cross-multiply when solving linear equations with fractions. This involves multiplying the numerator of one fraction by the denominator of the other fraction and setting them equal to each other. However, this method can often result in larger, more complicated fractions, so using the steps mentioned in question 2 may be more efficient.

4. What if the linear equation has multiple variables and fractions?

In this case, you can use the same steps as mentioned in question 2. However, you may need to use substitution or elimination methods to solve for the variables. This involves isolating one variable and substituting its value into the other equations until you are left with one equation and one variable to solve.

5. How can I check my answer when solving linear equations with fractions?

To check your answer, you can substitute the value you found for the variable back into the original equation and see if it makes the equation true. You can also graph the equation to see if the point of intersection between the two lines corresponds to the solution you found.

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