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Have to admit, I forgot how to do simple 9th grade algebra problems. I need to know how to do this stuff for an exam tomorrow. I am doing recurrences and the only thing I need to do is find $a, b$, and $c$ from the following

$$\begin{align*} a + b + c &= 3\\ 4a + 2b+ c&= 6\\ 9a + 3b + c& = 13 \end{align*}$$

Can someone refresh my memory on how to solve for $a, b$ and $c$ ?

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1  
Recall substitution, Gaussian Elimination/Row-Reduced-Echelon-Form, inverses? –  Amzoti Apr 11 '13 at 5:21
1  
See the example of Gaussian Elimination here, now you only need to substitute your numbers. –  sonystarmap Apr 11 '13 at 5:25

3 Answers 3

You have three unknowns and three equations, start by solving for one variable in one equation say "a" from the first one, then plug it into the second one and solve for another variable, for example:

(first equation):
$a = 3 - b - c$

(plug into second):
$4(3 - b - c) + 2b + c = 6$

$12 - 4b - 4c + 2b + c = 6$

$-2b - 3c = -6$

(solve for another variable):
$c = (-6 - 2b) / -3$

(use the results above and plug into third equation):
$9(3 - b - ((-6 - 2b) / -3))+3b+((-6 - 2b) / -3)=13$ from this equation you can solve for "b"

Continue this until you solve for all variables.

And so you can check your calculations:
$a = 2$
$b = -3$
$c = 4$

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Just use Gaussian elimination.

$a+b+c=3$

$4a+2b+c=6$

$9a+3b+c=13$

Subtracting equation 1 from equation 2 and equation 3:

$a+b+c=3$

$3a+b+0c=3$

$8a+2b+0c=10$

Subtract 2 times equation 2 from equation 3.

$a+b+c=3$

$3a+b+0c=3$

$2a+0b+0c=4$

We can see that $a=2$ Substitute this into the other equations

$2+b+c=3$

$3(2)+b+0c=3$

$a=2$

We can now see $b=-3$ Substitute this into equation 1.

$2+(-3)+c=3$

$b=-3$

$a=2$

Which shows that $c=4$

Testing these by putting them into the original equations.

$2+-3+4=3$

$4(2)+2(-3)+4=6$

$9(2)+3(-3)+4=13$

And we see they solve the system.

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You could represent your system as a single equation using matrices:

$$ \left[ \begin{matrix} 1 & 1 & 1\\ 4 & 2 & 1\\ 9 & 3 & 1 \end{matrix} \right] \left[ \begin{matrix} a\\ b\\ c \end{matrix} \right] = \left[ \begin{matrix} 3\\ 6\\ 13 \end{matrix} \right] $$

To isolate the matrix of unknowns, multiply on the left by the inverse of the matrix of coefficients:

$$ \left[ \begin{matrix} 1 & 1 & 1\\ 4 & 2 & 1\\ 9 & 3 & 1 \end{matrix} \right]^{-1} \left[ \begin{matrix} 1 & 1 & 1\\ 4 & 2 & 1\\ 9 & 3 & 1 \end{matrix} \right] \left[ \begin{matrix} a\\ b\\ c \end{matrix} \right] = \left[ \begin{matrix} 1 & 1 & 1\\ 4 & 2 & 1\\ 9 & 3 & 1 \end{matrix} \right]^{-1} \left[ \begin{matrix} 3\\ 6\\ 13 \end{matrix} \right] $$

which simplifies to

$$ \left[ \begin{matrix} a\\ b\\ c \end{matrix} \right] = \left[ \begin{matrix} 2\\ -3\\ 4 \end{matrix} \right] $$

That is, $a = 2$, $b = -3$, and $c = 4$.

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