Question

Solve the following inequation , write the solution set and represent it on the number line :
-3(x - 7) ≥ 15 - 7x > x + 1/3 . x ∈ R.

Answer 1

Answer:

{x ∈ R : -1.5 ≤ x ≤ 2}

Step-by-step explanation:

Given :

=> -3(x - 7) ≥ 15 - 7x > x + 1/3 . x ∈ R.

To find :

Solution set and represent it on the number line.

Solution :

=> -3(x - 7) ≥ 15 - 7x > x + 1/3 . x ∈ R

>> -3(x - 7) ≥ 15 - 7x (1)

>> 15 - 7x > (x + 1)/3 (2)

=> -3x + 21 ≥ 15 - 7x

=> 4x ≥ -6

=> x ≥ -3/2

=> -3/2 ≤ x

=> 15 - 7x > (x + 1)/3

=> 45 - 21x > x + 1

=> 45 - 1 > x + 21x

=> 44 > 22x

=> 2 > x

=> x < 2

>> -3/2 ≤ x < 2, x ∈ R

>> -1/5 ≤ x 2, x ∈ R

∴ Solution set:

=> {x ∈ R : -1.5 ≤ x ≤ 2}

Answer 2

$\large\underline{\sf{Solution-}}$

Given linear inequality is

$\red{\rm :\longmapsto\: - 3(x - 7) \geqslant 15 - 7x > x + \dfrac{1}{3} }$

Consider,

$\red {\rm :\longmapsto\: - 3(x - 7) \geqslant 15 - 7x }$

${\rm :\longmapsto\: - 3x + 21 \geqslant 15 - 7x }$

${\rm :\longmapsto\: - 3x + 7x \geqslant 15 - 21 }$

$\rm :\longmapsto\:4x \geqslant - 6$

$\bf\implies \:x \geqslant - \dfrac{3}{2} - - - (1)$

Now, Consider

$\red{\rm :\longmapsto\: 15 - 7x > x + \dfrac{1}{3} }$

${\rm :\longmapsto\: - 7x - x > \dfrac{1}{3} - 15}$

${\rm :\longmapsto\: - 8x > \dfrac{1 - 45}{3}}$

${\rm :\longmapsto\: - 8x > - \dfrac{44}{3}}$

We know,

$\boxed{ \bf{ \: x \geqslant y\bf\implies \: - x \leqslant - y}}$

$\bf :\implies\:x < \dfrac{11}{6} - - - (2)$

From equation (1) and (2), we concluded that

$\bf\implies \: - \dfrac{3}{2} \leqslant x < \dfrac{11}{6}$

$\bf\implies \:x \: \in \: \bigg[ - \: \dfrac{3}{2}, \: \dfrac{11}{6}\bigg)$

Additional Information :-

$\boxed{ \bf{ \: x > y\bf\implies \: - x < - y}}$

$\boxed{ \bf{ \: x < y\bf\implies \: - x > - y}}$

$\boxed{ \bf{ \: x < - y\bf\implies \: - x > y}}$

$\boxed{ \bf{ \: x > - y\bf\implies \: - x < y}}$

$\boxed{ \bf{ \: x \geqslant y\bf\implies \: - x \leqslant - y}}$

$\boxed{ \bf{ \: x \leqslant y\bf\implies \: - x \geqslant - y}}$

$\boxed{ \bf{ \: - x \leqslant y\bf\implies \: x \geqslant - y}}$

$\boxed{ \bf{ \: - x \geqslant y\bf\implies \: x \leqslant - y}}$