Solve the inequation and represent the solution set on the number line.
$- 3 + x \leq \frac{8 x}{3} + 2 \leq \frac{14}{3} + 2 x, w h e r e x ϵ I$

Open in App

Solution

$- 3 + x \leq \frac{8 x}{3} + 2 \leq \frac{14}{3} + 2 x, w h e r e x ϵ I - 3 + x \leq \frac{8 x}{3} + 2; \frac{8 x}{3} + 2 \leq \frac{14}{3} + 2 x \Rightarrow \frac{8 x}{3} - x \geq - 3 - 2; \frac{8 x}{3} - 2 x \leq \frac{14}{3} - 2 \Rightarrow \frac{5 x}{3} \geq - 5; \frac{2 x}{3} \leq \frac{8}{3} \Rightarrow 5 x \geq - 15; 2 x \leq 8 \Rightarrow x \geq - 3; x \leq 4 ∴ - 3 \leq x \leq 4 S o l u t i o n s e t = {- 3, - 2, - 1, - 0, 1, 2, 3, 4} .$

Suggest Corrections

Similar questions

Solve the inequation 5(x-2) > 4 (x + 3) -24 and represent it solution on a number line. Given the replacement set is {-4, -3, -2, -1, 0,1,2,3,4}.

- 3 + x \leq \frac{8 x}{3} + 2 \leq \frac{14}{3} + 2 x

x ϵ I

- 3 + x \leq \frac{8 x}{3} + 2 \leq \frac{14}{3} + 2 x, w h e r e x ϵ I

2 y - 3 < y + 1 \leq 4 y + 7; y ϵ R

- 3 + x \leq \frac{8 x}{3} + 2 \leq \frac{14}{3} + 2 x, w h e r e x ϵ I

2 y - 3 < y + 1 \leq 4 y + 7; y ϵ R

2 x - 5 \leq 5 x + 4 < 11, w h e r e x ϵ I

- 3 < - \frac{1}{2} - \frac{2 x}{3} \leq \frac{5}{6}, x ϵ R

Join BYJU'S Learning Program