TITLE: Solve the system of linear inequalities with parameters
QUESTION [5 upvotes]: Solve the system of inequalities
\begin{cases}
0\leq \phantom{-2\;}x+2\,y-3\,b+3\,a \leq 2 \\
0\leq -2\,x-3\,y+6\,b\phantom{\;+3a\;\,} \leq 1 \\[4pt]
0 \leq x \leq 1 \\   
0 \leq y \leq 2 \\[4pt]
0\leq a \leq 1 \\
0\leq b \leq 1 \tag{*}
\end{cases}
Here $x,y$ are unknown variables and $a,b$ are parameters.
My attempt. By adding the inequalities with some coeficients I separated the variables and get the simple system
\begin{cases}
0 \leq y+6a \leq 5,\\
0 \leq -x+9a+3b\leq 8.\\ \tag{**}
\end{cases}
and I am able to solve it. But the solutions of the last system  are not solution of the initial system!
Maple and wolframAlpha cant solve the system.
Any help?
P.S.1 For $a=\frac{63}{100}$ and $b=\frac{59}{100}$  ( as at G Cab's picture  below ) Maple gives the solutions
\begin{gather*}
 \left\{ x=1,{\frac{9}{50}}\leq y,y\leq {\frac{11}{25}} \right\} , \left\{ x=-3/2\,y+{\frac{127}{100}},{\frac{9}{50}}<y,y<{\frac{11}{25}} \right\} ,  \left\{ {\frac{9}{50}}<y,x<1,y<{\frac{11}{25}},-3/2\,y+{
\frac{127}{100}}<x \right\} , \left\{ y={\frac{11}{25}},{\frac{61}{100
}}\leq x,x<1 \right\} , \left\{ x=-3/2\,y+{\frac{127}{100}},{\frac{11}
{25}}<y,y<{\frac{127}{150}} \right\} ,  \left\{ {\frac{11}{25}}<y,x<-2
\,y+{\frac{47}{25}},y<{\frac{127}{150}},-3/2\,y+{\frac{127}{100}}<x
 \right\} , \left\{ x=-2\,y+{\frac{47}{25}},{\frac{11}{25}}<y,y<{\frac
{127}{150}} \right\} , \left\{ x=0,{\frac{127}{150}}\leq y,y\leq {
\frac{47}{50}} \right\} , \left\{ y={\frac{127}{150}},x\leq {\frac{14}
{75}},0<x \right\} , \left\{ 0<x,{\frac{127}{150}}<y,x<-2\,y+{\frac{47
}{25}},y<{\frac{47}{50}} \right\} , \\ \left\{ x=-2\,y+{\frac{47}{25}},{
\frac{127}{150}}<y,y<{\frac{47}{50}} \right\}
\end{gather*}
P.P.S. I am interested in finding not the whole set of solutions but in finding at least one solution from the area, if it exists,  but in terms $a$ and $b$, for example, $x=(a+b)/2, y=(2a-b)/10$ is a solution.

REPLY [2 votes]: 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 $\quad a=\dfrac{14}{17},\quad b=\dfrac14.$