System of equations

[troublemaker676]

I have to solve this system of equations for the three variables and my answers keep comming out wrong,

I'm eliminating c first,

1) 5a-b+3c=5--->x2--> 10a-2b+6c=10
2) 2a+7b-2c=5--->x3--> 6a+21b-6c=15 ----> *16a+19b=35, added equations
3) 4a-5b-7c=-65

2) 2a+7b-2c=5--->x7---> 14a+49b-14c=35
3) 4a-5b-7c=-65-->x2--> 8a-10b-14c=-130----> *6a+59b=165, subtracted equations

Then i brought down my two equations:

*16a+19b=35--->x3---> 48a+472b=1320
*6a+59b=165--->x8---> -48a+57b=105

I eliminated a and solved for b

I end up getting 415b=1215, which comes out to a really ugly fraction and i know there is no fractions in the answer. What did i do wrong?

 

[stapel]

I have to solve this system of equations...

Which system? You've posted some work (that I can't follow -- sorry), but what is the actual system? And what does "x2", "x3", etc, mean?

Thank you.

Eliz.

 

[troublemaker676]

1) 5a-b+3c=5
2) 2a+7b-2c=5
3) 4a-5b-7c=-65

This is the original system that i need to solve, "x3", "x2" means i multiplied the equations by either 2 or 3 to eliminate one of the variables. ex.

5a-b+3c=5--->x2--> 10a-2b+6c=10
2a+7b-2c=5--->x3--> 6a+21b-6c=15

The c's cancel out leaving me with just two variables.


I took the first two equations and eliminated c, then i took the last two equations and elminated c as well, so i was left with:

16a+19b=35
6a+59b=165

Then i decided to eliminate a and solve for b, this is where i get a weird fraction answer. I hope this helps to clear it up.

 

[Denis]

1) 5a-b+3c=5--->x2--> 10a-2b+6c=10
2) 2a+7b-2c=5--->x3--> 6a+21b-6c=15 ----> *16a+19b=35, added equations

Check your addition