Question

Solve the inequation:$\frac{5-2x}{3}$ $\le$ $\frac{x}{6}-5$, where x $\in$ N.

• A. {8, 9, 10 ……..}
• B. {1, 2, 3, 4 , 5, 6, 7 ……..}
• C. {-8, -9, -10 ……..}
• D. {0, 1, 2 ……..}

Answer 1

The correct option is A

{8, 9, 10 ……..}

$\frac{(5-2x)}{3}\le \frac{x}{6}-5$

Multiplying both sides by 6, we get:

$6\left[\frac{(5-2x)}{3}\right]\le 6[\frac{x}{6}-5]$

or $2(5-2x)\le x-30$

or $10-4x-10\le x-30-10$

or $-4x\le x-40$

or $-4x-x\le x-40-x$

or $-5x\le -40$

Dividing both sides by -5, as the number is negative, so the sign of inequality is reversed.

or $\frac{-5x}{-5}\ge -\frac{40}{-5}$

or $x\ge 8$

Hence, the solution set is {8, 9, 10 ……..}.