Question

if 6x<2-3x and x-7<3x what range of values of x satisfy both inequalities​

Answer 1

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

Given that

$\rm :\longmapsto\:6x < 2 - 3x - - - (1)$

and

$\rm :\longmapsto\:x - 7 < 3x - - - (2)$

Now,

Consider,

$\rm :\longmapsto\:6x < 2 - 3x$

☆ On adding 3x in each term, we get

$\rm :\longmapsto\:6x + 3x< 2 - 3x + 3x$

$\rm :\longmapsto\:9x < 2$

☆ On dividing by 9, each term, we get

$\bf\implies \:x < \dfrac{2}{9} - - - (3)$

Now,

Consider

$\rm :\longmapsto\:x - 7 < 3x$

☆ On Subtracting 3x in each term, we get

$\rm :\longmapsto\:x - 7 - 3x < 3x - 3x$

$\rm :\longmapsto\: - 2x - 7 < 0$

☆ On adding 7 in each term, we get

$\rm :\longmapsto\: - 2x - 7 + 7 < 0 + 7$

$\rm :\longmapsto\: - 2x < 7$

☆ On dividing each term by - 2, we get

$\bf :\longmapsto\:x > - \: \dfrac{7}{2} - - - (4)$

☆ From (3) and (4), we concluded that

$\rm :\longmapsto\:x \in \: \bigg( - \dfrac{7}{2} ,\dfrac{2}{9} \bigg)$

Additional Information :-

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

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

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

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

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

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

$\boxed{ \green{ \tt \: \dfrac{x}{y} > 0 \implies \: x > 0, \: y > 0 \: or \: x < 0, \:y < 0}}$

$\boxed{ \green{ \tt \: \dfrac{x}{y} < 0 \implies \: x > 0, \: y < 0 \: or \: x > 0, \:y < 0}}$