Question

Find all the positive integer valued solutions $(x, y, z)$ of the systems of inequalities
$ \begin{cases} 3x+2y-z=4, & \quad \text{} \\ 2x-y+2z=6, & \quad \text{} \\ x+y+x<7. & \quad \text{} \end{cases} $
Determine how many such triplets exist?

• A. $0$
• B. $1$
• C. $3$
• D. None of these

Answer 1

Answer: (B)

Given equations are
$3x + 2y - z = 4, …(i)$
$2x - y + 2z = 6 …(ii)$
$x + y + z < 7. …(iii)$
By $(i) + 2 × (ii),$
$7x + 3z = 16,$
$\therefore x = 1, z = 3.$
From $(i), y = 2x + 2z - 6 = 2$. Since $x = 1,$
$y = 2, z = 3$ satisfy the inequality $(iii)$, the solution is $x = 1, y = 2, z = 3.$