# Solve three equations with three unknowns [closed]

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?

• now i have the solution from a book but i dont know how, i want someone explain the method for me – J-rone Jun 28 '17 at 20:11
• @J-rone The solution is ... Any permutation of $1,2,3$ is a solution as well e.g. $a=2,b=3,c=1$ since the equations are symmetric in $a,b,c$. – dxiv Jun 28 '17 at 20:21
• The upvotes are baffling considering this post has no context. – StubbornAtom Jun 29 '17 at 12:41
• @StubbornAtom Not at all this is a 'Hot Network Question' hence the upvotes. – kingW3 Jun 29 '17 at 13:21

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 $\{1,2,3\}$.

• Nice, @dxiv (out of votes for the next 3 and a half hours, but I'll make a point of keeping the tab open!) This, I think, best establishes how to move from the given system of equations, to arrive at the solutions, as requested by the asker. – Namaste Jun 28 '17 at 20:39
• @amWhy Thanks. As a side note, the problem with questions like this one, posted without any context, is that it gives no clues about OP's background, or what kind of answer they are in fact looking for. – dxiv Jun 28 '17 at 20:46
• @dxiv They're looking for how to solve that system of equations. How complicated can that be? – LastStar007 Jun 28 '17 at 21:15
• @Max As written in the answer, $a^3-6 a^2+11a - 6 = 0$ has $a=1$ as a root, which is pretty obvious by inspection, or could otherwise more systematically be determined using the rational root theorem. Therefore $a^3-6 a^2+11a - 6$ must be divisible by $a-1$. – dxiv Jun 29 '17 at 5:50
• Thank you for your explanations, that clears it up for me! – Max Jun 29 '17 at 5:51

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 $\{a,b,c\}=\{1,2,3\}$.

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.

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.

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: $$(a,b,c)=(1,2,3); (1,3,2); (2,1,3); (2,3,1); (3,1,2); (3,2,1).$$

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 $3$.