Question

In the algebraic inequality $9\ge -3x\ge -3$, how many integral values does $x$ have?

• A. $4$
• B. $3$
• C. $5$

Answer 1

The correct option is C $5$

Given algebraic inequality : $9\ge -3x\ge -3$
Step 1: First, we will solve $9\ge -3x$
On dividing both sides of the inequality by $-3$, we get
$\Rightarrow \frac{9}{-3}\le \frac{-3x}{-3}\Rightarrow -3\le x$

Step 2: Now, we will solve $-3x\ge -3$
On dividing both sides of the inequality by $-3$, we get
$\Rightarrow \frac{-3x}{-3}\le \frac{-3}{-3}\Rightarrow x\le 1$

$\therefore x$ is greater than or equal to $-3$ and less than or equal to $1$
$\Rightarrow -3\le x\le 1$
Integers between $-3\&1$ including $-3\&1$ are $-3,-2,-1,0,1$.