Solving for the range of lengths of a irregular quadrilateral given two sides of equal (unspecified) length and one fixed side. Given an irregular Quadrilateral with two opposing sides of length called constant D, one side fixed as 74 units, and another with length equal to 10x-4, find the range of possible values for x. 
The problem is question is merely for algebra one, it is unlikely the techniques involved are anything more than simple geometry. However, I find myself stuck with it.
Methods tried:
I have split the Quadrilateral into two triangles that share a hypotenuse H and one leg of length D
Using Triangle inequalities I can gather two equations for each of the two triangles: 
D > H - 74
And 
D < H + (10x + 4)
Solving the two leaves me with a single bound for X, that is: 
X > -7 
Obviously this is not helpful, as the length of one side of the triangle must be greater than 0. When I realized this, I instead solved for
10x - 4 > 0 
x > 4/10
But how might I find the upper bound on X?
 A: Consider the side length of $74$ being "fixed" in space, with the vertices of the two sides of length $D$ away from the $74$ side moving apart as much as they can, such as indicated in the diagram below.

This shows the upper limit for the length of the $10x - 4$ side would occur where the sides of length $D$, $74$ and $D$ are almost in a straight line, so the sum of their lengths would just barely be more than $10x - 4$. In that case, you get
$$\begin{equation}\begin{aligned}
D + 74 + D & \gt 10x - 4 \\
2D + 78 & \gt 10x \\
0.2(D) + 7.8 & \gt x
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
As for the lower limit, consider instead the vertices of the two sides of length $D$ away from the side length of $74$ moving together until they can touch, which occurs if $2D \ge 74$ and in which case you get $0.4 \lt x$ which you determined yourself, or until they basically almost reach the side of length $74$. In that latter case, as indicated in the diagram below,

the minimum length of the remaining side must be more than the difference of lengths, i.e., $74 - 2D$. Thus, consider 
$$E = \max(0, 74 - 2D) \tag{2}\label{eq2A}$$
The more general lower limit would then come from
$$\begin{equation}\begin{aligned}
10x - 4 & \gt E \\
10x & \gt E + 4 \\
x & \gt 0.1(E) + 0.4
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Thus, you then have
$$0.1(E) + 0.4 \lt x \lt 0.2(D) + 7.8 \tag{4}\label{eq4A}$$
as the possible range for $x$.
