Question

solve for x, |x+1|+|x|>3​

Answer 1

Answer:

$x \in( - \infty \:, \: 1) \bigcup(1 \: , \infty )$

Step-by-step explanation:

Make cases

case 1 :

x < -1

Both mod will open with negative sign

$- (x + 1) - x - 3 > 0 \\ - 2x - 4 > 0 \\ - 2x > 4 \\ \: \: \: \: x < 2$

Now take intersection with case we got

$x \in \: \: ( - \infty \: \: 2)\bigcap \: ( - \infty \: ,- 1) \: = ( - \infty \: ., - 1)$

case 2

-1 < x < 0

Only | x | open with negative sign

$x + 1 - x > 3 \\ 1 > 3$

Which is false . So we got no solution from here.

case 3

x > 0

Both mod opens with positive sign

$x + 1 + x > 3 \\ x > 1$

Taking intersection with our case we got

$x \in \: (1 \: \:,\: \infty )$

So final result