Solve the system of linear inequalities with parameters

When you have a system of (double) inequalities like this $$ \left\{ \matrix{ a \le x \le b \hfill \cr c \le x \le d \hfill \cr} \right. $$ you may think that each one represents a segment on the $x$ axis, with the system standing for AND, i.e. $$ \eqalign{ & \left\{ \matrix{ x \in \left[ {a,b} \right] \hfill \cr x \in \left[ {c,d} \right] \hfill \cr} \right.\quad \Rightarrow \quad x \in \left( {\left[ {a,b} \right] \cap \left[ {c,d} \right]} \right)\quad \Rightarrow \cr & \Rightarrow \quad x \in \left[ {\max (a,c),\min \left( {b,d} \right)} \right] \cr} $$

Therefore in your case we can do some manipulation as follows $$ \eqalign{ & \left\{ \matrix{ 0 \le x + 2y + 3a - 3b \le 2 \hfill \cr 0 \le - 2x - 3y + 6b \le 1 \hfill \cr 0 \le x \le 1 \hfill \cr 0 \le y \le 2 \hfill \cr 0 \le a \le 1 \hfill \cr 0 \le b \le 1 \hfill \cr} \right. \Rightarrow \cr & \Rightarrow \left\{ \matrix{ 3\left( {b - a} \right) \le x + 2y \le 3\left( {b - a} \right) + 2 \hfill \cr 6b - 1 \le 2x + 3y \le 6b \hfill \cr 0 \le x \le 1 \hfill \cr 0 \le y \le 2 \hfill \cr 0 \le a \le 1 \hfill \cr 0 \le b \le 1 \hfill \cr} \right. \Rightarrow \cr & \Rightarrow \left\{ \matrix{ 0 \le a \le 1 \hfill \cr 0 \le b \le 1 \hfill \cr 0 \le y \le 2 \hfill \cr 3\left( {b - a} \right) \le x \le 3\left( {b - a} \right) + 2 - 2y \hfill \cr 3b - 1/2 - 3/2y \le x \le 6b - 3/2y \hfill \cr 0 \le x \le 1 \hfill \cr} \right. \Rightarrow \cr & \Rightarrow \left\{ \matrix{ 0 \le a \le 1 \hfill \cr 0 \le b \le 1 \hfill \cr 0 \le y \le 2 \hfill \cr m = \max \left( {3\left( {b - a} \right),3b - 1/2 - 3/2y,0} \right) \hfill \cr n = \min \left( {3\left( {b - a} \right) + 2 - 2y,\;6b - 3/2y,\;1} \right) \hfill \cr m \le x \le n \hfill \cr} \right. \cr} $$

where at the third step we chose to isolate the $x$, but of course we could have done that with $y$ in this case obtaining $$ \left\{ \matrix{ 0 \le a \le 1 \hfill \cr 0 \le b \le 1 \hfill \cr 0 \le x \le 1 \hfill \cr m = \max \left( {3/2\left( {b - a} \right) - x/2 2b - 1/3 - 2/3x 0} \right) \hfill \cr n = \min \left( {3/2\left( {b - a} \right) + 1 - x/2 \;2b - 2/3x \;2} \right) \hfill \cr m \le y \le n \hfill \cr} \right. $$

So, once fixed $a, \; b, \; y$ within the allowed range, we can finish to compute $x$ in the first case, or viceversa in the second version.

The system sketched in Geogebra gives

note in reply to your comment

As you can see from the sketch, the solutions (when they exist) will in general define a 2D area.
Referring to the case depicted, once fixed $a$ and $b$, you can describe the area by having the $y$ to span the allowed range $[0,2]$ and consequently determine $x$ to be within two bounds, necessarily depending from $y$.
There is no possibility to express the bounds on $x$ and $y$ independently from each other.

example with $a=0.63 ,\; b=0.59$

$$ \begin{array}{l} \left\{ \begin{array}{l} a = 0.63 \\ b = 0.59 \\ 0 \le y \le 2 \\ m = \max \left( { - 0.12,1.27 - 3/2y,0} \right) \\ n = \min \left( {1.88 - 2y,\;3.54 - 3/2y,\;1} \right) \\ m \le x \le n \\ \end{array} \right.\;\; \Rightarrow \\ \Rightarrow \left\{ \begin{array}{l} 0 \le y \le 2 \\ \begin{array}{*{20}c} {1.27 - 3/2y \le x \le 1} \hfill & {\left| {\;0 \le y < 0.44} \right.} \hfill \\ {1.27 - 3/2y \le x \le 1.88 - 2y} \hfill & {\left| {\;0.44 \le y < 2.54/3} \right.} \hfill \\ {0 \le x \le 1.88 - 2y} \hfill & {\left| {\;2.54/3 \le y < 0.94} \right.} \hfill \\ {0 \le x \le 1.88 - 2y\; \to \;\emptyset } \hfill & {\left| {\;0.94 \le y \le 2} \right.} \hfill \\ \end{array} \\ \end{array} \right.\; \Rightarrow \\ \Rightarrow \left\{ {\begin{array}{*{20}c} {1.27 - 3/2y \le x \le 1} \hfill & {\left| {\;0.18 \le y < 0.44} \right.} \hfill \\ {1.27 - 3/2y \le x \le 1.88 - 2y} \hfill & {\left| {\;0.44 \le y < 2.54/3} \right.} \hfill \\ {0 \le x \le 1.88 - 2y} \hfill & {\left| {\;2.54/3 \le y < 0.94} \right.} \hfill \\ \end{array}} \right. \\ \end{array} $$

The second version instead gives the more simple result $$ \left\{ \matrix{ a = 0.63 \hfill \cr b = 0.59 \hfill \cr 0 \le x \le 1 \hfill \cr m = \max \left( { - 0.06 - x/2,1.18 - 1/3 - 2/3x,0} \right) = \hfill \cr = 1.18 - 1/3 - 2/3x \hfill \cr n = \min \left( {0.94 - x/2,\;1.18 - 2/3x,\;2} \right) = \hfill \cr = 0.94 - x/2 \hfill \cr m \le y \le n \hfill \cr} \right. $$

addendum

I do not catch properly your requirement, but in any case I will introduce a more geometric approach to the problem that might offer a different view of the solutions.

Each double inequality represent a stripe comprised between two parallel lines with constant separation. The two stripes overlap to define a parallelogram, which is just translated while maintaining its shape unaltered.

The coordinates of the four vertices are $$ \begin{array}{c|cccc} {} & & {Vsi} & {Vss} & {Vii} & {Vis} \\ \hline x & & {9a + 3b - 8} & {9a + 3b - 6} & {9a + 3b - 2} & {9a + 3b} \\ y & & { - 6a + 5} & { - 6a + 4} & { - 6a + 1} & { - 6a} \\ \end{array} $$ Now, until the x of $Vsi$ is greater than $1$ the whole parallelogram $P$ will be out of the rectangle $R = [0,1] \times [0,2]$.
Same if the y of $Vsi$ is below zero. So to have solutions it shall be $$ \eqalign{ & \left\{ \matrix{ 0 \le b \le 2 \hfill \cr 0 \le a \le 1 \hfill \cr 9a + 3b - 8 \le 1 \hfill \cr 0 \le - 6a + 5 \hfill \cr} \right.\quad \Rightarrow \quad \left\{ \matrix{ 0 \le b \le 2 \hfill \cr 0 \le a \le 1 - b/3 \hfill \cr a \le 5/6 \hfill \cr} \right.\quad \Rightarrow \cr & \Rightarrow \quad \left\{ \matrix{ 0 \le b \le 1/2\; \wedge \;0 \le a \le 5/6 \hfill \cr 1/2 < b \le 2\; \wedge \;0 \le a \le 1 - b/3 \hfill \cr} \right. \cr} $$ Note that the above is a necessary condition yet not sufficient. That's because when the upper vertex is in the second quadrant we have still to impose that $P$ intersects $R$, which is as complicated as to use the previous min/max conditions.

Let us consider the system in the form of \begin{cases} 6b-1\le 2x+3y \le 6b\\ 3b-3a \le x+2y \le 2+3b-3a\\ x,y \in[0,1]\tag1 \end{cases} over the set of the possible pairs $(a,b)\in[0,1]^2.$

$$\color{blue}{\mathbf{Case\ 1.\quad a-b >\dfrac23.}}$$

The system $(1)$ does not have solutions.

$$\color{blue}{\mathbf{Case\,2.\quad 0\le a \le \min\left[\frac{2+3b}3,1\right].}}\tag2$$

$\color{blue}{\mathbf{Case\,2.1.\quad b\in \bigg[0,\dfrac16\bigg],\quad a\in\bigg[0,b\bigg].}}$

The first equation of the system in the form of \begin{cases} 0 \le 2x+3y \le 6b\\ 3b-3a\le x+2y \le 2+3b-3a\tag{3.1} \end{cases} over the first quadrant defines the triangle with the vertice $\quad (0,0),\quad (3b,0),\quad (0,2b).$

The second equation over the first quadrant defines the trapezoid with the vertice

$(3b-3a, 0),\quad (2+3b-3a,0),\quad (0, \frac{2+3b-3a}2),\quad(0, \frac{3b-3a}2).$

Since

• $\ 0 \le 3b-3a\le 3b \le 2+3b-3a,$

• $\ 0 \le \frac{3b-3a}2 \le 2b \le \frac{2+3b-3a}2,$

then the solution is the simplex with the vertice $(3b-3a, 0),\quad (3b,0),\quad (0,2b),\quad (0, \frac{3b-3a}2).$

Analytically, $$\bigg(x\in\bigg[0,3b\bigg]\bigg)\wedge\bigg(y\in\bigg[\max\left(\frac{3b-3a-x}2,0\right),\frac{6b-2x}3\bigg]\bigg).\tag{4.1}$$

Solution for $\quad a=\dfrac1{10},\quad b=\dfrac18.$

$\color{blue}{\mathbf{Case\,2.2.\quad b\in \bigg[0,\dfrac16\bigg],\quad a\in\bigg[b,\dfrac{2+3b}3\bigg].}}$

The first equation of the system in the form of \begin{cases} 0 \le 2x+3y \le 6b\\ 0\le x+2y \le 2+3b-3a\tag{3.2} \end{cases} over the first quadrant defines the triangle with the vertice $\quad (0,0),\quad (3b,0),\quad (0,2b).$

The second equation over the first quadrant defines the triangle with the vertice

$(0,0),\quad (2+3b-3a,0),\quad (0, \frac{2+3b-3a}2).$

Since

• equality $2b = \frac{2+3b-3a}2$ takes place if $a=\frac{2-b}3,$
• equality $3b = 2+3b-3a$ takes place if $a=\frac{2}3,$

then the next cases should be considered.

$$\color{green}{\mathbf{Case\,2.2.1.\quad b\in \left[0,\dfrac16\right],\quad a\in\bigg[b,\dfrac{2-b}3\bigg].}}$$

Solution is the triangle with the vertice $\quad (0,0),\quad (3b,0),\quad (0,2b).$

Analytically, $$\bigg(x\in\bigg[0,3b\bigg]\bigg)\wedge\bigg(y\in\bigg[0, \frac{6b-2x}3\bigg]\bigg).\tag{4.2.1}$$

Solution for $\quad a\in\bigg[\dfrac1{8},\dfrac58\bigg],\quad b=\dfrac18.$

$$\color{green}{\mathbf{Case\,2.2.2.\quad b\in \left[0,\dfrac16\right],\quad a\in\bigg[\dfrac{2-b}3,\dfrac23\bigg].}}$$

The lines $2x+3y=6b$ and $x+2y = 2+3b-3a$ have intersection at the point $(x_i,y_i) = (9a+3b-6, 4-6a).$

Solution is the simplex with the vertice $\quad (0,0),\quad (3b,0),\quad (9a+3b-6, 4-6a),\quad (0,\frac{2+3b-3a}2).$

Analytically, $$\bigg(x\in\bigg[0,3b\bigg]\bigg)\wedge\bigg(y\in\bigg[0, \min\left(\frac{2+3b-3a-x}2,\frac{6b-2x}3\right)\bigg]\bigg).\tag{4.2.2}$$

Solution for $\quad a = \dfrac{9}{14},\quad b=\dfrac18.$

$$\color{green}{\mathbf{Case\,2.2.3.\quad b\in \left[0,\dfrac16\right],\quad a\in\bigg[\dfrac23,\dfrac{2+3b}3\bigg].}}$$

Solution is the triangle with the vertice $\quad (0,0),\quad (2+3b-3a,0),\quad (0,\frac{2+3b-3a}2).$

Analytically, $$\bigg(x\in\bigg[0,2+3b-3a\bigg]\bigg)\wedge\bigg(y\in\bigg[0, \frac{2+3b-3a-x}2\bigg]\bigg).\tag{4.2.3}$$

Solution for $\quad a = \dfrac{17}{24},\quad b=\dfrac18.$

$\color{blue}{\mathbf{Case\,2.3.\quad b\in \bigg[\dfrac16,\dfrac13\bigg],\quad a\in\bigg[0,b\bigg].}}$

The first equation of the system in the form of \begin{cases} 6b-1 \le 2x+3y \le 6b\\ 3b-3a\le x+2y \le 2+3b-3a\tag{3.3} \end{cases} over the first quadrant defines the trapezoid with the vertice $\quad (\frac{6b-1}2,0),\quad (3b,0),\quad (0,2b),\quad (0,\frac{6b-1}3).$

The second equation over the first quadrant defines the trapezoid with the vertice

$(3b-3a, 0),\quad (2+3b-3a,0),\quad (0, \frac12(2+3b-3a)),\quad (0, \frac12(3b-3a)).$

Since

• equality $\frac{6b-1}3 = \frac{3b-3a}2$ takes place if $a=\frac{2-3b}9,$
• equality $\frac{6b-1}2 = 3b-3a$ takes place if $a=\frac16,$

then the next cases should be considered.

$$\color{green}{\mathbf{Case\,2.3.1.\quad b\in \left[\dfrac16,\dfrac13\right],\quad a\in\bigg[0,\dfrac{2-3b}9\bigg].}}$$

Solution is the simplex with the vertice

$(3b-3a,0),\quad (3b, 0),\quad (0,2b),\quad (0, \frac{3b-3a}2).$

Analytically, $$\bigg(x\in\bigg[0,3b\bigg]\bigg)\wedge\bigg(y\in\bigg[\max\left(\frac{3b-3a-x}2,0\right),\frac{6b-2x}3\bigg]\bigg).\tag{4.3.1}$$

Solution for $\quad a=\dfrac1{12},\quad b=\dfrac14.$

$$\color{green}{\mathbf{Case\,2.3.2.\quad b\in \left[\dfrac16,\dfrac13\right],\quad a\in\bigg[\dfrac{2-3b}9,\dfrac16\bigg].}}$$

The lines $2x+3y=6b-1$ and $x+2y = 3b-3a$ have intersection at the point $(x_i,y_i) = (9a+3b-2,1-6a).$

Solution is the simplex with the vertice

$(3b-3a,0),\quad (3b, 0),\quad (0,2b),\quad (0, \frac{6b-1}3),\quad (9a+3b-2,1-6a).$

Analytically, $${\small\bigg(x\in\bigg[0,3b\bigg]\bigg)\wedge\bigg(y\in\bigg[\max\left(\frac{6b-1-2x}3, \frac{3b-3a-x}2,0\right),\frac{6b-2x}3\bigg]\bigg)}.\tag{4.3.2}$$

Solution for $\quad a=\dfrac3{19},\quad b=\dfrac14.$

$$\color{green}{\mathbf{Case\,2.3.3.\quad b\in \left[\dfrac16,\dfrac13\right],\quad a\in\bigg[\dfrac16,b\bigg].}}$$

Solution is the trapezoid with the vertice

$(3b,0),\quad (3b-3a, 0),\quad (0, \frac{3b-3a}2),\quad (0,2b).$

Analytically, $$\bigg(x\in\bigg[0,3b\bigg]\bigg)\wedge\bigg(y\in\bigg[\max\left(\frac{6b-1-2x}3,0\right),\frac{6b-2x}3\bigg]\bigg).\tag{4.3.3}$$

Solution for $\quad a=\dfrac15,\quad b=\dfrac14.$

$\color{blue}{\mathbf{Case\,2.4.\quad b\in \left[\dfrac16,\dfrac13\right],\quad a\in\bigg[b,b+\dfrac23\bigg].}}$

The first equation of the system in the form of \begin{cases} 6b-1 \le 2x+3y \le 6b\\ 0 \le x+2y \le 2+3b-3a\tag{3.4} \end{cases}

over the first quadrant defines the trapezoid with the vertice $\quad (\frac{6b-1}2,0),\quad (3b,0),\quad (0,2b),\quad (0,\frac{6b-1}3).$

The second equation over the first quadrant defines the triangle with the vertice

$(0,0),\quad (2+3b-3a,0),\quad (0, \frac{2+3b-3a}2).$

Since

• equality $2b = \frac{2+3b-3a}2$ takes place if $a=\frac{2-b}3,$

• equlaity $3b = 2+3b-3a$ takes place if $a = \frac23,$

• equaity $\frac{6b-1}3 = \frac{2+3b-3a}2$ takes place if $a=\frac{8-3b}9,$

• equaity $\frac{6b-1}2 = 2+3b-3a$ takes place if $a=\frac56,$

then the next cases should be considered.

$$\color{green}{\mathbf{Case\,2.4.1.\quad b\in \left[\dfrac16,\dfrac13\right],\quad a\in\bigg[0,\dfrac{2-b}3\bigg].}}$$

Solution is the trapezoid from the paragraph $2.3.3$ above.

$$\color{green}{\mathbf{Case\,2.4.2.\quad b\in \left[\dfrac16,\dfrac13\right],\quad a\in\bigg[\dfrac{2-b}3,\dfrac23 \bigg].}}$$

The lines $2x+3y=6b$ and $x+2y = 2+3b-3a$ have intersection at the point $(x_i,y_i) = (9a+3b-6,4-6a).$

Solution is the simplex with the vertice

$(\frac{6b-1}2,0),\quad (3b, 0),\quad (9a+3b-6,4-6a),\quad (0,\frac{2+3b-3a}2),\quad (0, \frac{6b-1}3).$

Analytically, $${\small\bigg(x\in\bigg[0,3b\bigg]\bigg)\wedge\bigg(y\in\bigg[\max\left(\frac{6b-1-2x}3,0\right),\min\left(\frac{2+3b-3a-x}2,\frac{6b-2x}3\right)\bigg]\bigg)}.\tag{4.4.2}$$

Solution for $\quad a=\dfrac58,\quad b=\dfrac14.$

$$\color{green}{\mathbf{Case\,2.4.3.\quad b\in \left[\dfrac16,\dfrac13\right],\quad a\in\bigg[\dfrac23,\dfrac{8-3b}9 \bigg].}}$$

Solution is the simplex with the vertice

$(\frac{6b-1}2,0),\quad (2+3b-3a, 0),\quad (0,\frac{2+3b-3a}2),\quad (0, \frac{6b-1}3).$

Analytically, $${\small\bigg(x\in\bigg[0,2+3b-3a\bigg]\bigg)\wedge\bigg(y\in\bigg[\max\left(\frac{6b-1-2x}3,0\right),\frac{2+3b-3a-x}2\bigg]\bigg)}.\tag{4.4.3}$$

Solution for $\quad a=\dfrac34,\quad b=\dfrac14.$

$$\color{green}{\mathbf{Case\,2.4.4.\quad b\in \left[\dfrac16,\dfrac13\right],\quad a\in\bigg[\dfrac{8-3b}9,\dfrac56 \bigg].}}$$

The lines $2x+3y=6b-1$ and $x+2y = 2+3b-3a$ have intersection at the point $(x_i,y_i) = (9a+3b-8,5-6a).$

Solution is the triangle with the vertice

$(\frac{6b-1}2,0),\quad (2+3b-3a, 0),\quad (9a+3b-8,5-6a).$

Analytically, $${\small (x\in[9a+3b-8,2+3b-3a])\wedge\bigg(y\in\bigg[\max\left(\frac{6b-1-2x}3,0\right),\frac{2+3b-3a-x}2\bigg]\bigg)}.\tag{4.4.4}$$

Solution for $\quad a=\dfrac{14}{17},\quad b=\dfrac14.$