Solve the following inequation- -2 - p - 3- - 5 - p- varepsilon - I

Question

Toppr Admin · Accepted Answer

Given- -2 - p - 3- -gt- 5 -160- -2p - 6 -gt- 5 -160- -2p -gt- 5 - 6 -160- -2p -gt- 11 -160-We get-160- -p -gt- -dfrac-11-2- -160- p -lt- -dfrac-11-2- -160-As -160- p-160- -varepsilon - I - -160-Therefore - solution set-160- - - -left - - -8 - -7 - -6 -right - -160-

Solve the following inequation:
$$ -2 ( p + 3) > 5 , p\, \varepsilon \, I $$

Given
$$ -2 ( p + 3) > 5 $$
$$ -2p - 6 > 5 $$
$$ -2p > 5 + 6 $$
$$ -2p > 11 $$
We get,
$$ -p > \dfrac{11}{2} $$

$$ p < \dfrac{-11}{2} $$
As $$ p\, \varepsilon \, I $$ ,
Therefore , solution set $$ = \left \{ ..... -8 , -7 , -6 \right \} $$

Solve the following inequation:$$ -2 ( p + 3) > 5 , p\, \varepsilon \, I $$

Given$$ -2 ( p + 3) > 5 $$ $$ -2p - 6 > 5 $$ $$ -2p > 5 + 6 $$ $$ -2p > 11 $$ We get, $$ -p > \dfrac{11}{2} $$ $$ p < \dfrac{-11}{2} $$ As $$ p\, \varepsilon \, I $$ , Therefore , solution set $$ = \left \{ ..... -8 , -7 , -6 \right \} $$

Solve the following inequation:
$$ -2 ( p + 3) > 5 , p\, \varepsilon \, I $$

Given
$$ -2 ( p + 3) > 5 $$
$$ -2p - 6 > 5 $$
$$ -2p > 5 + 6 $$
$$ -2p > 11 $$
We get,
$$ -p > \dfrac{11}{2} $$

$$ p < \dfrac{-11}{2} $$
As $$ p\, \varepsilon \, I $$ ,
Therefore , solution set $$ = \left \{ ..... -8 , -7 , -6 \right \} $$