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September 21st, 2014, 05:58 AM  #1 
Newbie Joined: Sep 2014 From: Toronto Posts: 7 Thanks: 0  Systems of 3 Linear Equations solutions
i need help with this question, thanx all question: Consider the system of linear equations: x+y+z=6 3x+2y+z=10 mx+2y+z=n For what values of m and n does the system of equations have: 1) no solutions? 2) a unique solution? 3) infinitely many solutions? 
September 21st, 2014, 08:34 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra 
Equation 3 less equation 2 $$(m3)x = n10$$ So $m \ne 3$ has at one solution for $x$ (you need to check that you get unique solutions for $y$ and $z$). $m = 3, n=10$ implies that any value of $x$ will work. Again this can be verified (and fine tuned) in the system. $m=3, n \ne 10$ has no solutions. Last edited by v8archie; September 21st, 2014 at 08:42 AM. 
September 21st, 2014, 08:34 AM  #3 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038  
September 21st, 2014, 08:55 AM  #4  
Newbie Joined: Sep 2014 From: Toronto Posts: 7 Thanks: 0  Quote:
solve x+y+z=6 for z: Z=6xy substitute into the other two equations: so, 3x+2y+z=10; 2x+y=4 and, 5x+2y+z=12; 4x+y=6 solve by eliminating y out of the above two equations: x=1 then by substituting x into 2x+y=4, you get y= 2, then by substituting both x and y into z=6xy you get the value of z which is 3. so: x=1, y=2 and z=3 i know how to do the basics but its hard for a 10th grader to these kind of questions when ur teacher doesnt explain how to do it  
September 21st, 2014, 10:31 AM  #5 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038  
September 27th, 2014, 03:55 PM  #6 
Newbie Joined: Sep 2014 From: Toronto Posts: 7 Thanks: 0  Oh I'm sorry I didn't realize this was a grammar forum..... I'll remember that the next time I go on a MATH forum. And no need to get so mad. I didn't realize saying ur instead of YOUR would mean so much. In the future if you happen to get annoyed with someone's questions just ignore it. This is a math forum, I'm pretty sure that means you come to get help, not to get sour and annoyed remarks from people. You don't want to help? Then mind YOUR own business.

September 27th, 2014, 05:18 PM  #7 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 
Well, if that's the way you're taking it, so be it. Good luck. 

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equations, linear, solutions, systems 
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