Question

help me in this question guys!!

$solve$
$\frac{ |x + 3| + x}{x + 2} > 1$
​

Answer 1

$<b> <u> Heya! </b> </u>$

____________________________________________________________

$<b> <u> <i> Linear Inequalities</i> </b> </u>$

→ Given Equation :

$= > \frac{ |x + 3| + x }{x + 2} > 1 \\ \\ = > \frac{ |x + 3| + x }{x + 2} - 1 > 0 \\ \\ = > \frac{ |x + 3| + x - x - 2 }{x + 2} > 0 \\ \\ = > \frac{ |x + 3| - 2 }{x + 2} > 0$

•°• $<u> We have two cases now </u>$

→ $<b> Case 1 : When x ≥ -3 </b>$

$= > \: in \: this \: case = |x + 3| = x + 3 \\ \\ = > \frac{x + 3 - 2}{x + 2} > 0 \\ \\ = > \frac{x + 1}{x + 2} > 0$

°•° For this equation , solution set is -

$<b> x € ( - ∞ , -2 ) U ( -1 , ∞ ) </b>$

But here x≥ -3 so the solution set is

$<b> <u> [ -3 , -2 ) U ( -1 , ∞ ) </u> </b>$

________________________________________

→ $<b> Case 2 : When x < -3 </b>$

$= > \frac{ - (x + 3) - 2}{x + 2} > 0 \\ \\ = > \frac{ - (x + 5)}{x + 2} > 0 \\ \\ = > \frac{x + 5}{x + 2} < 0$

°•° For this equation , solution set is -

$<b> x € ( -5 , -3 ) </b>$ [ by using x <-3 ]

$<i><u>from Case 1 and 2 we get the final solution set : </i></u>$

=>$<b> Solution = ( -5 , -2 ) U ( -1 , ∞ ) ✔</b>$

___________________________________________________________

Answer 2

Step-by-step explanation:

Answer:x=-5,x=2

Step-by-step explanation:

(x+1)(x+2)+(x-2)(x-1)/(x-1)(x+2)=3

x^2+2x+x+2+x^2-x-2x+2/x^2+2x-x-2=3

2x^2+3x-3x+4=3x^2+3x-6

2x^2+4=3x^2+3x-6

0=3x^2-2x^2+3x-10

x^2+3x-10=0

x^2+(5-2)x-10=0

x^2+5x-2x-10=0

x(x+5)-2(x+5)=0

(x+5)(x-2)

x=-5

x=2