Question

Find the quadrant which contains no solution of the given system of inequalities $x+3y\le 12$ and $2x-3y\le -3$ .

• A. Quadrant I
• B. Quadrant IV
• C. Quadrant I and IV
• D. Quadrant III and IV

Answer 1

Answer: (B)

Quadrant IV

Given, $x+3y\le 12$

Multiply by $2$ on both sides, we get

$2x+6y\le 24$........(1)

$2x-3y\le -3$.........(2)

Subtract (1) by (2), we get

$-9y=\le -27$

$\Rightarrow y\le 3$

Put the value of $y$ in equation (1), we get

$x+3\times 3\le 12$

$\Rightarrow x\le 12-9=3$

Then the solution is $(3,3)$ which is not in Quadrant IV.