Question

Find the maximum value of $x$ that satisfies the following inequation:
$-3(x-7)\ge 15-7x>\frac{x+1}{3};x\in \text{W}.$

• A. 1

Answer 1

The correct option is A 1

Given: $-3(x-7)\ge 15-7x>\frac{x+1}{3}$

let's separate the given inequation in two inequations as shown below:

$\begin{array}{c}-3(x-7)\ge 15-7x\\ -3x+21\ge 15-7x\\ 4x\ge -6\\ x\ge \frac{-6}{4}\\ x\ge -1.5\end{array}|\begin{array}{c}15-7x>\frac{x+1}{3}\\ 15-\frac{1}{3}>\frac{x}{3}+7x\\ \frac{45-1}{3}>\frac{x+21x}{3}\\ 44>22x\\ 2>x\end{array}$

$\therefore$ $-1.5\le x<2$

The same solution set can be represented on the number line as shown below:

The whole numbers lying between -1.5 and 2 are: 0 and 1. Hence, the maximum value of $x$ that satisfies the inequation is 1.