Question

$find the range of values of x that satisfy the following pair of inequality 2x-3 \leqslant 5 and 7-6x \leqslant -3$
​

Answer 1

Given :-

$\rm :\longmapsto\:2x - 3 \leqslant 5$

$\rm :\longmapsto\:7 - 6x \leqslant - 3$

To Find :-

• Range of values of x

Solution :-

Consider,

$\red{\rm :\longmapsto\:2x - 3 \leqslant 5}$

$\rm :\longmapsto\:2x \leqslant 5 + 3$

$\rm :\longmapsto\:2x \leqslant 8$

$\rm :\longmapsto\:x \leqslant 4$

$\red{\bf\implies \:x \in \: ( - \infty ,4]} - - - (1)$

Consider,

$\green{\bf :\longmapsto\:7 - 6x \leqslant - 3}$

$\rm :\longmapsto\: - 6x \leqslant - 3 - 7$

$\rm :\longmapsto\: - 6x \leqslant - 10$

$\rm :\longmapsto\:x \geqslant \dfrac{10}{6}$

$\rm :\longmapsto\:x \geqslant \dfrac{5}{3}$

$\green{\bf\implies \:x \in \: \bigg[\dfrac{5}{3}, \infty )} - - - - (2)$

Hence, Range of x satisfied both the inequalities

$\green{\bf\implies \:x \in \: \bigg[\dfrac{5}{3}, \infty ) \: \cap \: ( - \infty,4] }$

$\green{\bf\implies \:x \in \: \bigg[\dfrac{5}{3}, \: 4 \bigg] }$

Additional Information :-

$\boxed{ \sf \:a < - b \implies \: - a > b}$

$\boxed{ \sf \: - a < - b \implies \: a > b}$

$\boxed{ \sf \: a > - b \implies \: - a < b}$

$\boxed{ \sf \: - a > - b \implies \: a < b}$

$\boxed{ \sf \:a \leqslant - b \implies \: - a \geqslant b}$

$\boxed{ \sf \: - a \leqslant - b \implies \: a \geqslant b}$

$\boxed{ \sf \: a \geqslant - b \implies \: - a \leqslant b}$

$\boxed{ \sf \: - a \geqslant - b \implies \: a \leqslant b}$