1cat2ferrets Posted December 4, 2008 Share Posted December 4, 2008 in 3 variables!! My son and I are nearly in tears trying to figure this problem out. Very confusing, to say the least!!:confused: Here it is: 3x^2+4y+z=7 2y+z=3 -5x+3y+8z=-31 Thanks to anyone that can help us figure it out!! Quote Link to comment Share on other sites More sharing options...
NicksMama-Zack's Mama Too Posted December 4, 2008 Share Posted December 4, 2008 but here is a free website with some help... http://www.mathway.com/problem.aspx?p=algebra K Quote Link to comment Share on other sites More sharing options...
Jann in TX Posted December 4, 2008 Share Posted December 4, 2008 (edited) does the first equation have an x squared? Much easier to deal with if it doesn't! Also--has he been using the 'elimination' method or is he using matrices yet? Edited December 4, 2008 by Jann in TX Quote Link to comment Share on other sites More sharing options...
MyThreeSons Posted December 5, 2008 Share Posted December 5, 2008 Here it is: 3x^2+4y+z=7 2y+z=3 -5x+3y+8z=-31 Assuming you've mastered 2 equations with 2 unknowns, one way to tackle this problem would be to solve the middle equation for z: z = 3 - 2y Plug that expression in for z in the first and third equations, and you have 2 equations with 2 unknowns. Solve those for x and y, then go back to the middle one and solve for z. Note: having that x^2 term in the first equation definitely complicates things. HTH Quote Link to comment Share on other sites More sharing options...
annabanana1992 Posted December 5, 2008 Share Posted December 5, 2008 If these are supposed to be systems of linear equations why is there an x squared term? Quote Link to comment Share on other sites More sharing options...
1cat2ferrets Posted December 5, 2008 Author Share Posted December 5, 2008 Yes, Jann, there's an x squared in the first equation. We finally found out how to do it after a couple of hours. Thanks for your reply. Quote Link to comment Share on other sites More sharing options...
1cat2ferrets Posted December 5, 2008 Author Share Posted December 5, 2008 Thanks for your help. We finally found out how to do this problem after awhile. Quote Link to comment Share on other sites More sharing options...
1cat2ferrets Posted December 5, 2008 Author Share Posted December 5, 2008 That's what the problem stated and that's why we were confused, but figured it out after a couple of hours. Thanks anyway! Quote Link to comment Share on other sites More sharing options...
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