3 Variable Linear Equation using Elimination... HELP!!!

[burton4550]

Hi all, first post, and I firstly want to say, I'm not looking for answers, just wanting to learn how to do this correctly. I've been having all kinds of issues with it and could really use someone to let me know what I'm doing wrong? Ok so here is what I have so far.


Every time I try and solve this problem I keep getting fractions. I can't figure out how to solve it and just get whole numbers.


Here's the problem. 3 equations, 3 variables, need to solve using the elimination method. Any help would be much appreciated.

x -2 +z = 6

2x +y -3 = 3

x -3y +3z = 10



I took -2 and multiplied it by the top row

Then added this to the second row.

-2x +4 -2z =-12

2x +y -3 = 3

I canceled out the 2x

5y -5z = -9

I set this aside and did another equations to get an equation with variables y and z only.

I took the top one and multiplied it by -1. Then I added it to the 3rd row. The x's cancel out.

-x 2 -z = -6

x -3y +3z = 10



-1y +2z = 4

Now I have 2 equations with only 2 variables.

I multiplied the bottom line by 1 to cancel out the y's. Then I added them.

5y -5z = -9

-1y +2z = 4



-5y +2z = 4

5y -5z = -9

-3z=-5

Then I divided -3 from both sides.

then I get z =5/3



Then I plug this in to the equation with only 2 variables to solve for x.

x -2 + 5/3 = 6

-2x + 5/3 = 6

-2x + 5/3 = 3/10

Divide -2 from both sides

x = 6/100 simplify

x=3/50



I'm going to stop here cause I already know it's wrong and there's no point to keep going. Please help with this, it's seriously driving me crazy!

 

[stapel]

Every time I try and solve this problem I keep getting fractions. I can't figure out how to solve it and just get whole numbers.

If the solution values actually are fractions, then there is no possible way to "just get whole numbers" as a correct solution. Why do you think that the solution must be whole numbers?

...need to solve using the elimination method.

x -2 +z = 6
2x +y -3 = 3
x -3y +3z = 10

From what you've posted (below), I'm going to guess that the original system was actually as follows:

. . . . .\(\displaystyle \begin{array}{rrrrrcr}x&-&2y&+&z&=&6\\2x&+&y&-&3x&=&3\\x&-&3y&+&3z&=&10\end{array}\)

I took -2 and multiplied it by the top row

Then added this to the second row.

This means that you added -2R₁ + R₂:

. . . . .\(\displaystyle -2x\, +\, 4y\, -\, 2z\, =\, -12\)
. . . . .\(\displaystyle +2x\, +\, 1y\, -\, 3z\, =\, +3\)

...to get a new R₂:

. . . . .\(\displaystyle 0x\, +\, 5y\, -\, 5z\, =\, -9\)

I set this aside and did another equations to get an equation with variables y and z only.

I took the top one and multiplied it by -1. Then I added it to the 3rd row. The x's cancel out.

This means that you added -1R₁ + R₃:

. . . . .\(\displaystyle -1x\, +\, 2y\, -\, 1z\, =\, -6\)
. . . . .\(\displaystyle +1x\, -\, 3y\, +\, 3z\, =\, 10\)

...to get a new R₃:

. . . . .\(\displaystyle 0x\, -\, 1y\, +\, 2z\, =\, 4\)

Now I have 2 equations with only 2 variables.

Sort of. The first row still exists; you're just ignoring it for the moment. The full system is now:

. . . . .\(\displaystyle \begin{array}{rrrrrcr}x&-&2y&+&z&=&6\\0x&+&5y&-&5z&=&-9\\0x&-&1y&+&2z&=&4\end{array}\)

I multiplied the bottom line by [5] to cancel out the y's. Then I added them.

This means that you multiplied the new R₃ by 5, adding 5R₃ + (new) R₂:

. . . . .\(\displaystyle 0x\, -\, 5y\, +\, 10z\, =\, 20\)
. . . . .\(\displaystyle 0x\, +\, 5y\, -\, 5z\, =\, -9\)

...to get a new R₂:

. . . . .\(\displaystyle 0x\, +\, 0y\, +\, 5z\, =\, 11\)

The system is now:

. . . . .\(\displaystyle \begin{array}{rrrrrcr}1x&-&2y&+&z&=&6\\0x&+&0y&+&5z&=&11\\0x&-&1y&+&2z&=&4\end{array}\)

Something went wrong in your computations; it looks like maybe you only multiplied part of the equation by 5...? Anyway, you can continue your working from the above. But the new second row makes it clear that, since 5z = 11, there is no way that this can have a whole-number solution.

 

[burton4550]

Thank you for the help on this one. I was actually able to figure it out. I did have the first equation written wrong which is why I kept getting the wrong answers for the variables.

I left out the y like you mentioned in that first equation