Question

Solve the linear inequations and represent the solution set on the number line.
(a) $9\ge 7x-5\ge 5x-11,x\in R$

(b) $-2\le \frac{1}{2}-\frac{2x}{3}\le 1\frac{5}{6},x\in N$

• A. (a) $[-3,1]$
  (b) $\frac{15}{4}\ge x\ge -2$
• B. (a) $[0,2]$
  (b) $\frac{15}{4}\ge x\ge -2$
• C. (a) $[-3,2]$
  (b) $\frac{15}{4}\ge x\ge -2$
• D. (a) $[0,5]$
  (b) $\frac{15}{4}\ge x\ge -2$

Answer 1

The correct option is C (a) $[-3,2]$

(b) $\frac{15}{4}\ge x\ge -2$
Part $\left(a\right)$

$9\ge 7x-5\ge 5x-11$

Now,

$9\ge 7x-5$

$14\ge 7x$

$2\ge x$

Now,

$7x-5\ge 5x-11$

$2x\ge -6$

$x\ge -3$

Therefore,

$x\in [-3,2]$

Part $\left(b\right)$

$-2\le \frac{1}{2}-\frac{2x}{3}\le 1\frac{5}{6}$

Now,

$-\frac{5}{2}\le -\frac{2x}{3}$

$-\frac{15}{6}\le -\frac{4x}{6}$

$\frac{15}{4}\ge x$

Now,

$\frac{1}{2}-\frac{2x}{3}\le 1\frac{5}{6}$

$\frac{1}{2}-\frac{2x}{3}\le \frac{11}{6}$

$-\frac{2x}{3}\le \frac{8}{6}$

$-\frac{2x}{3}\le \frac{4}{3}$

$-2x\le 4$

$-x\le 2$

$x\ge -2$

Therefore,

$\frac{15}{4}\ge x\ge -2$

Hence, this is the answer.