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Consider:

$a + b = 5$

$2a + b + c = 4$

$a - b - c = 5$

I like to use substituiton for solving systems of equations, so I firstly look at equation 1 and solve for $a$

$a + b = 5$

$a = 5 - b$

I substitute this into equation 2

$2(5-b) + b + c = 4$

$10 -2b + b + c = 4$

$-b = 4 - c - 10$

$b = -4 + c + 10$

What next? Am I on the right track? Can I use substitution method for any simultaneous equation?

xvzz
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  • Yes, go on : you eliminated $a$ from the first, eliminate now $b$ from the second and you are left with one equation where you will find $c$. – Claude Leibovici Jun 12 '14 at 13:45
  • Have you been exposed to Gaussian elimination? This is possibly the best (non-numerical) way to approach such systems: http://en.wikipedia.org/wiki/Gaussian_elimination – Andreas Caranti Jun 12 '14 at 13:53

2 Answers2

1

You have the right concept, and we can see from three equations and three unknowns that these variables are solvable (taking into account equations are linearly independent)

equation 1: $\ a+b=5$

equation 2:$\ 2a+b+c= 4$

equation 3:$\ a-b-c =5$

lets solve for b first:

from equation 1 and 3, we say

$a=5-b$ and $\ c = a-b-5$

Now we plug these two 'new' equations into equation 2:

$\ 2(5-b) + b + ((5-b) - b -5) = 4 $

simplifying for b gives that $ b=2$.

Now taking that b=2, we can sub that back into equation 1 and solve for a and once we have a we can use a and b to solve for c. Which I'm sure you'll be able to solve for it.

JLL
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Substitution is not the best way to go about this kind of problem. Use reduction instead. For example if we add the second and third equations together we get $$3a=9$$ so we know that $a=3$ immediately. The first then gives us that $b=2$ and finally $c=-4$. Much easier and hardly any work.

Rene Schipperus
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