# Creative Algebra Net Problem Solving Question

I came across a problem that I found pretty tedious and difficult to answer and I would appreciate any views or solutions for this question.

The diagram shows the net of a cube. On each face there is an integer: $$1$$, $$2011$$,$$1207$$, $$x$$, $$y$$ and $$z$$. If each of the numbers $$1207$$, $$x$$, $$y$$ and $$z$$ equals the average of the numbers written on the four faces of the cube adjacent to it, find the value of $$x$$. The net is provided for the question below.

Using a rather obvious approach, I began to list out all the averages for $$1207$$, $$x$$, $$y$$, and $$z$$.

I ended up with these 4 equations (not simplified):

$$1207$$=$$(x$$+$$y$$+$$2011$$+$$z)$$/$$4$$

$$x$$=$$(1207$$+$$y$$+$$1$$+$$z$$)$$/4$$

$$y$$=$$(x$$+$$2011$$+$$1207$$+$$1)$$/$$4$$

$$z$$=$$(2011$$+$$1207$$+$$1$$+$$x)$$/$$4$$

I am now stuck at this step and I am unable to proceed and correctly evaluate the values for $$x$$, $$y$$ and $$z$$. I originally thought that some subsitution and rearranging was requried but falied to solve it in the end.

I would appreciate any help that would allow me to solve this problem. However if you are able to solve the above set of equations, that would definitely be the most helpful to me. Nevertheless, I would accept any type of solution to this problem. Thank you very much!!

• The tag net is reserved for a different concept, as net has a technical meaning in point set topology (generalizing that of a convergenec of a sequence). I am quite sure that the choice of tag was inappropriate here. May be graph-theory is what you want, because graphs can be used to describe the relation of adjacency (here adjacency) of the faces of a cube. Commented Jul 18 at 6:49
• @JyrkiLahtonen: I wouldn't use graph theory here. This is a linear algebra problem disguised as a graph. Commented Jul 18 at 18:15

Since you have $$4$$ linear equations in $$3$$ unknowns, it's an overdetermined system and, thus, potentially inconsistent, but this answer shows this isn't the case. First, to avoid dealing with fractions, multiply both sides of each of your equations by $$4$$ to get

\begin{aligned} 4\times 1207 & = x + y + 2011 + z \\ 4x & = 1207 + y + 1 + z \\ 4y & = x + 2011 + 1207 + 1 \\ 4z & = 2011 + 1207 + 1 + x \end{aligned}\tag{1}\label{eq1A}

Next, the second equation minus the first one gives

\begin{aligned} 4x - 4\times 1207 & = -x + (1207 - 2011) + 1 \\ 5x & = 5\times 1207 - 2011 + 1 \\ x & = 805 \end{aligned}\tag{2}\label{eq2A}

Although it's not requested, note the RHS of the third and fourth equations in \eqref{eq1A} are the same, i.e., so $$y = z$$. Thus, using this and the result of \eqref{eq2A} in the first equation in \eqref{eq1A} leads to

\begin{aligned} 4\times 1207 & = 805 + 2011 + 2z \\ 4\times 1207 - 805 - 2011 & = 2z \\ z & = 1006 \end{aligned}\tag{3}\label{eq3A}

Since $$y = z$$, we get the solution of $$(x, y, z) = (805, 1006, 1006)$$, with these values being consistent with the remaining $$3$$ equations in \eqref{eq1A}.