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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
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Re: System of 3 equations with 3 variables

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 Originally Posted by 500lbgorilla I tried to look up a way to do this but I am more confused than when i started.

 January 21st, 2009, 08:59 AM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 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. $Eq1:\mbox{ }20=400A+20B+C$ $Eq2:\mbox{ }105=1600A+40B+C$ $Eq3:\mbox{ }295=3600A+60B+C$ Multiply Eq1 by 2 to get $Eq4:\mbox{ }40=800A+40B+2C$ Multiply Eq1 by 3 to get $Eq5:\mbox{ }60=1200A+60B+3C$ Subtract Eq4 from Eq2 to get $Eq6:\mbox{ }65=800A-C$ Subtract Eq5 from Eq3 to get $Eq7:\mbox{ }235=2400A-2C$ Now, with Eq6 and Eq7, we have a system of two equations in two variables. Multiply Eq6 by 2 to get $130=1600A-2C$ Subtract this from Eq7 and we get $105=800A$ So A = 21/160, sub this into Eq6 getting $65=800\left(\frac{21}{160}\right)-C$ Solve for C, C = 40. Since A = 21/160 and C = 40, it follows that B = -29/8 so our quadratic equation is $y=\frac{21}{160}x^2-\frac{29}{8}x+40$
 January 24th, 2009, 07:59 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,926 Thanks: 2205 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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