Solving for multiple variables given a couple of equations? I have a panel 1200 pixels wide, and am filling in smaller subpanels to fill the length. Each sub-panel is of a different color ($p$ = purple, $g$ = green, etc). It's for a navigation bar on a website, each subpanel corresponds to a link to another page.
In one situation, there's three purple subpanels, a yellow subpanel, two blue subpanels, one each of green and red subpanels; in another situation, there's 3 purple subpanels, a yellow ($e$ for clarification) subpanel, two more purple subpanels, and a green subpanel. The width of each subpanel (irrespective of situation) must add up to 1200px. 
Lastly, $e$ is always the middle panel, with some panels to the left or right. The combined widths of the left must equal the combined widths of the right in both situations.
Since all subpanel widths must add up to 1200px irrespective of situation, two equations can be made:
$$\begin{align}
3p + e + 2b + g + r &= 1200 \\
3p + e + 2p + g &= 1200
\end{align}$$
Since $e$ is in the center of the panel, subpanels to the left and right of $e$ in the above equations must equal. This gives me two more:
$$\begin{align}
3p &= 2b + g + r \\
3p &= 2p + g
\end{align}$$
I think I can derive two more equations, but I'm not sure their relevance. Since $e$ is in the middle ($e$'s midpoint is in the middle of the panel), the left and right halves each contain half of $e$:
$$\begin{align}
3p + \frac{1}{2}e &= \frac{1}{2}e + 2b + g + r \\
3p + \frac{1}{2}e &= \frac{1}{2}e + 2p + g
\end{align}$$
Alternatively:
$$\begin{align}
3p + \frac{1}{2}e &= 600 \\
\frac{1}{2}e + 2b + g + r &= 600 \\
\frac{1}{2}e + 2p + g &= 600
\end{align}$$
I want to find any solution given a couple of rules:
$$\begin{align}
p &> 150\\
b &> 100 \\
e &> 90 \\
g &> 80 \\
r &> 60
\end{align}$$
As long as any solution fits those minimum values, things are kosher.
Forming a matrix and bringing to RREF yields three free variables, which is why I'm stumped.
The illustrate the first situation, $3p + e + 2b + g + r = 1200$:

 A: By subtracting the first from the second you get $2p=2b+r.$ Then you can express each of $e,r$ in terms of the other three variables:
$$r=2p-2b,\\ e=1200-5p-g.$$
There are thus three degrees of freedom in the equation system. Are there other 
It seems from your added material that the yellow sections lasbelled $e$ are to be located at the center of each long panel, making extra equations. call these the "balance" requirements. Then you do have from your second balance requirement that $3p=2p+e$ which implies that in fact $p=e.$ This brings down the number of variables to four. You have from the first balance equation that $3p=2b+g+r.$ So now there are in effect three equations in four varibles, there would then expectedly be only one free parameter.
Keeping only the variables $p,b,e,r$ in that order, since the second balance equation implies $e=p,$ we arrive at the three equations
$$4p+2b+e+r=1200,\\ 6p+e=1200,\\ 2p-2b-r=0,$$ the last coming from the first balance equation. When this system is row reduced, one gets that $r=0$ follows, and also $p=b=200-e/6.$ Likely something is off in the set-up, if $r=0$ is implied, since that would mean width 0 for the red strips.
Added: OOPS I entered the matrix to be row reduced incorrectly. As it turns out, we now still have $g=p$ from one of the balance equations, while the 3-rowed matrix row reduces to one in which the last row is all zeroes. The other two rows let us read off a parametric solution giving $p,b$ each in terms of $e,r.$ Namely $p=200-e/6,\ b=200-e/6-r/2.$ 
We can now use the inequality requirements on the variables to obtain a region in the $e,r$ plane. We have $200-e/6>150.$ [Since $g=p$ there is no need to use also the constraint $g>80$ since it follows from $p>150.$] Then we also have $200-e/6-r/2>100.$ Then in the $e,r$ plane there are the inequalities $e>90$ and $r>60.$ I haven't done the graph, but it is now simple to simultaneously graph these four inequalities and get a region in the $e,r$ plane, and if that region is nonempty you have found solutions.
Check this, but I did the graph and found it has a simple description: $90<e<300$ along with $60<r<200-e/6,$ which is nonempty and any pair $e,r$ satisfying these leads back to a solution which you called "kosher". (To get integer solutions one needs $6|e$ and $2|r.$)
