Solve three equations with three unknowns Solve the system:
$$\begin{cases}a+b+c=6\\ab+ac+bc=11\\abc=6\end{cases}$$
The solution is:
$a=1,b=2,c=3$
How can I solve it?
 A: For class work it is likely that the roots are integers, so I would just try them.  There are not many factorizations of $6$ and $1,2,3$ should jump out.  Then just try it and you are done.  
The routine approach is substitution.  Write the first as $a=6-b-c$ and plug that into the other two.  Solve the second for $b$ and you have one (messy) equation in $c$.  The rational root theorem will work here.
A: You should post your attempts at the question.
Try rearranging some things to fit them together.
$$\begin{cases}a+b+c=6\\ab+ac+bc=11\\abc=6\end{cases}$$

$$a+b+c=6$$
$$a+c=6-b$$

$$ab+ac+bc=11$$
$$b(a+c)+ac=11$$

$$b(a+c)+ac=11=b(6-b)+ac=11$$
$$ac=\frac{6}{b}$$
$$b(6-b)+ac=11=b(6-b)+\frac{6}{b}=11$$
now you have an equation dependent only upon b.
Rearrange it to find b.  Then substitute the answer for b into the other equations and rearrange and substitute to solve for a or c.  Then substitute that in with b and you can find whichever variable is left.
A: By method of substitution:
$$\begin{cases} b+c=6-a \\ a(6-a)+\frac{6}{a}=11 \\ bc=\frac{6}{a}\end{cases} \Rightarrow$$
$$a^3-6a^2+11a-6=0 \Rightarrow (a-1)(a-2)(a-3)=0 \Rightarrow a=1; 2; 3.$$
Substituting these and solving $$\begin{cases} b+c=6-a \\ bc=\frac{6}{a}\end{cases}$$
six solutions will be found:
$$(a,b,c)=(1,2,3); (1,3,2); (2,1,3); (2,3,1); (3,1,2); (3,2,1).$$
A: As an alternative to the other derivations, here's my approach.
In the second equation, move everything to the left side, factor $ac+bc$ into $(a+b)c$, multiply everything by $c$, replace $a+b$ with $6-c $ from the first equation, and $abc$ with $6$ from the third. Then you have $6-11c+(6-c)c^2 = 0$, which simplifies to $c^3-6c^2+11c-6 = 0$, which has roots $1, 2,$ and $3$.
A: Without Vieta's formulas, the polynomial can be derived the hard way by elimination. From the first and last equations:
$$
b+c=6-a \\
bc = 6 / a
$$
Substituting the above into the middle equation:
$$
11=a(b+c)+bc=a(6-a)+6/a \;\;\iff\;\; a^3-6 a^2+11a - 6 = 0
$$
By inspection, the latter equation has $a=1$ as a root, then factoring out $a-1$ gives:
$$a^3-6 a^2+11a - 6=(a-1)(a^2-5a+6)=(a-1)(a-2)(a-3)$$
Since the original system is symmetric in $a,b,c$ it follows that the solutions are $\{1,2,3\}$.
A: By using Vieta's formulas, the system is equivalent to say that $a,b,c$ are the roots of $x^3-6x^2+11x-6$, which factorizes as $(x-1)(x-2)(x-3)$. Hence $\{a,b,c\}=\{1,2,3\}$.
