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January 20th, 2009, 02:13 PM   #1
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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?
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January 20th, 2009, 02:23 PM   #2
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Re: System of 3 equations with 3 variables

Google: Solve three equations
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January 20th, 2009, 05:26 PM   #3
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Re: System of 3 equations with 3 variables

Quote:
Originally Posted by 500lbgorilla
I tried to look up a way to do this but I am more confused than when i started.
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January 21st, 2009, 08:59 AM   #4
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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

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January 24th, 2009, 07:59 AM   #5
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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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