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January 20th, 2009, 02:13 PM  #1 
Newbie Joined: Jan 2009 Posts: 7 Thanks: 0  System of 3 equations with 3 variables
Ok so here's the problem, and I tried to look up a way to do this but I am more confused than when i started. Find a quadratic function that fits the data on the right. The 3 points given are (20,20) (40,105) & (60,295) Using the points I can input into the quadratic model and get: 20= 400A+20B+C 105= 1600A+40B+C 295= 3600A+60B+C After this is where I'm lost and seriously need help. Can someone please help me work this and explain it in a way that makes some kind of sense? 
January 20th, 2009, 02:23 PM  #2 
Senior Member Joined: Jul 2008 Posts: 895 Thanks: 0  Re: System of 3 equations with 3 variables
Google: Solve three equations

January 20th, 2009, 05:26 PM  #3  
Newbie Joined: Jan 2009 Posts: 7 Thanks: 0  Re: System of 3 equations with 3 variables Quote:
 
January 21st, 2009, 08:59 AM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,929 Thanks: 1124 Math Focus: Elementary mathematics and beyond  Re: System of 3 equations with 3 variables
The idea is to multiply the given equations by a factor that produces a common (equal) term which we can then eliminate from the equations using subtraction. Multiply Eq1 by 2 to get Multiply Eq1 by 3 to get Subtract Eq4 from Eq2 to get Subtract Eq5 from Eq3 to get Now, with Eq6 and Eq7, we have a system of two equations in two variables. Multiply Eq6 by 2 to get Subtract this from Eq7 and we get So A = 21/160, sub this into Eq6 getting Solve for C, C = 40. Since A = 21/160 and C = 40, it follows that B = 29/8 so our quadratic equation is 
January 24th, 2009, 07:59 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,467 Thanks: 2038 
Eq3  Eq2 gives Eq4: 2000A + 20B = 190. Eq2  Eq1 gives Eq5: 1200A + 20B = 85. Eq4  Eq5 gives 800A = 105, so A = 21/160. Eq3  3*Eq5 gives C = 295  3(85) = 40. 5*Eq5  3*Eq4 gives 40B = 425  570, so B = 29/8. 

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