# Solving Linear System with inequalities

I have the following system:

\begin{align} b - x = 0 \\ a - 0.33b - 0.5x =0 \\ d - 0.33b = 0 \\ a - 0.33b + c = 0 \\ a + b + c + d + 2x = 1 \\ a + b + c + d - 8.8x \le 0 \\ a + b + c + d - 7.27x \ge 0 \end{align}

where a,b,c,d and x are unknown variables.

Is there any solution to this system and how to solve this system?

• What did you try? – wythagoras May 4 '15 at 9:31
• actually I am new to this Linear programming and linear systems, so I help me out. – Parvez Kazi May 4 '15 at 9:37
• What do you know about $b$ and $x$? – wythagoras May 4 '15 at 9:43
• As I said, a,b,c ,d and x are all unknown variables. – Parvez Kazi May 4 '15 at 9:48
• I will give you a hint. Is the 0.33 actually meant to be 1/3? – wythagoras May 4 '15 at 9:51

Since $$x=b$$ $$d=0.33b$$ $$a=0.33b+0.5b=0.83b$$ $$c=0.33b-0.83b=-0.5b$$ one has $$1=(0.83+1-0.5+0.33+2)b\Rightarrow b=\frac{1}{3.66}$$ Hence, $$a=\frac{0.83}{3.66},b=\frac{1}{3.66},c=-\frac{0.5}{3.66},d=\frac{0.33}{3.66},x=\frac{1}{3.66}.$$

However, $$a+b+c+d-7.27x=\frac{0.83+1-0.5+0.33-7.27}{3.66}=\frac{-5.61}{3.66}\color{red}{\lt} 0$$ so, the system has no solution.

• Well done, but what about the inequalities ? – Yves Daoust May 4 '15 at 11:12
• @YvesDaoust: Thank you for pointing it out. I should have written about them. – mathlove May 4 '15 at 11:20

Some hints:

1. $b-x=0$, and thus $b=x$.
2. Now we have $a-0.5x-0.33b=0$, thus $a-0.5x-0.33x=0$, thus $a-0.83x=0$, and thus $a=0.83x$.
3. What do we know of $d$ in the form of $b$?

No, this system has no solution.

The equations give a matrix

(%i) A;
[ 0    1     0  0   - 1  ]
[                        ]
[ 1  - 0.33  0  0  - 0.5 ]
[                        ]
(%o)                      [ 0  - 0.33  0  1    0   ]
[                        ]
[ 1  - 0.33  1  0    0   ]
[                        ]
[ 1    1     1  1    2   ]


with non-zero determinant $3.66$, so $A x = b$, with solution vector $b$

(%i) b;
[ 0 ]
[   ]
[ 0 ]
[   ]
(%o)                                 [ 0 ]
[   ]
[ 0 ]
[   ]
[ 1 ]


has exactly one solution $x$.

(%i) x;
[  0.22677595628415  ]
[                    ]
[  0.27322404371585  ]
[                    ]
(%o)                        [ - 0.13661202185792 ]
[                    ]
[ 0.090163934426229  ]
[                    ]
[  0.27322404371585  ]


Inserting this solution into the inequalities gives:

(%i) [1,1,1,1,-8.8] . x;
(%o)                        - 1.950819672131148
(%i) [1,1,1,1,-7.27] . x;
(%o)                        - 1.532786885245901


The first one ("$\le 0$") holds, the second one ("$\ge 0$") not.