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Simultaenous Equations : Problem

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by nightriders_leo » Mon Mar 16, 2009 11:30 am
There are three equation -- 1) a+b+c=18
2) 4a+2b+c=27
3) 9a+3b+c=38

if you substract equation(2) - equation(1) & equation(3) - equation(2)

4(a)+2(b)+(c)=27 9(a)+3(b)+c=38
1(a)+1(b)+(c)=18 4(a)+2(b)+c=27
---------------------- -------------------
3(a)+(b) = 9 5(a)+(b) = 11
---------------------- ---------------------

On Solving these two equation or say Substracting these equation

5(a)+(b)=11
3 (a)+(b)=9
-----------------
2(a) = 2 so (a)=1
-----------------

So now value of a = 1 substituting the value of a=1 in

equation 5(a)+(b)=11 so it will be 5+b = 11

so value of b = 6

same way substitute the value of a & B in equation(1)

a+b+c = 18 so 1+6+(c)=18

value of c=11

so here are the three values a=1 b= 6 c=11
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Saurabh
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Re: Simultaenous Equations : Problem

by Morgoth » Mon Mar 16, 2009 11:35 am
BlindVision wrote:a(1)^2 + b(1) + c = 18

a(2)^2 + b(2) + c = 27

a(3)^2 + b(3) + c = 38


I need help in understanding how the answers came to be...

a = 1, b = 6, c = 11

Thank you!


a + b + c = 18 -------I

4a + 2b + c = 18-----II

9a + 3b + c = 18------III


firstly subtract II & I

4a + 2b + c = 27
a + b + c = 18

3a + b = 9

simplify the III equation

9a + 3b + c = 38

3(3a + b) + c = 38

substitute the value of 3a + b

3*9 + c = 38

c = 38 - 27 = 11

We already know 3a+b=9

b = 9 - 3a

substitute the values in I equation

a+b+c = 18

a + 9 - 3a + 11 = 18

20 - 18 = 2a

a= 2/2 = 1

substitute the value of a in 3a + b = 9

3 + b = 9

b = 9-3 = 6

Hence, a = 1, b = 6 & c = 11

Hope this helps.
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by cramya » Mon Mar 16, 2009 2:51 pm
Welcome back Morgoth!

Hope u have been doing well buddy

Regards,
Cramya
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by Morgoth » Wed Mar 18, 2009 12:20 am
cramya wrote:Welcome back Morgoth!

Hope u have been doing well buddy

Regards,
Cramya
Thanks for the welcome buddy. Hope you are doing good as well. Nice to see that you are still enlightening Gmater's in their cause to beat the gmat.
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