System of inequalities with 3 variables

Question

I have a system of inequalities:

\begin{array}{c} a-b+c>0 \\ a+b+c<4 \\ 9a-3b+c<-5 \end{array}

Wolfram|Alpha says that \begin{array}{c} a < -\frac {1}{8} \end{array} But how can I solve this system of inequalities for this answer?

Answer 1

Since $a+c>b$ we get $2b<4$ so $b<2$.

Also $$9a-3b+c = (a+c)+8a-3b>8a-2b$$ so we get $$8a<2b-5<4-5=-1$$

and thus the claim.

Answer 2

Using the first equation, we can easily determine that :

$a+c>b\qquad\color{blue}{(1)}$

In the second equation, if we use $\,(1)\,$ we will reach the conclusion that $\,2b<4\,$ as, $\,b+b<a+b+c<4\,$.

Therefore $\;b<2\qquad\color{blue}{(2)}$

Now in the $\,3^\text{rd}$ equation we observe $8a-3b+b<8a-3b+a+c<-5\quad$ (using $\;(1)\;$)

So, $\;8a-2b<-5$

But, $\;b<2\quad(2)$

Therefore $\;8a-4<-5$

and hence $\;8a<-1\;.$

Answer 3

Solving the system

$$ \cases{ a-b+c=s_1^2\\ a+b+c-4=-s_2^2\\ 9a-3b+c+5=-s_3^2 } $$

we obtain the solutions

$$ \cases{ a=-\frac 18-\frac 18(2s_1^2+s_2^2+s_3^2)\\ b=2-\frac 12(s_1^2+s_2^2)\\ c=\frac 18(17+6s_1^2-3s_2^2+s_3^3) } $$

so we can conclude easily that

$$ a < -\frac 18, \ \ b < 2 $$