Question

$\Large{\underline{\underline{\bf{Question:-}}}}$
Solve the given inequality for real x: $\frac{x}{3} > \frac{x}{2} + 1$​

Answer 1

Answer:

Given :-

x/3 > (x/2)+1

To find :-

Solve the given inequality for real x ?

Solution :-

Given in-equation is x/3 > (x/2)+1

On subtracting (x/2) both sides then

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

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

LCM of 3 and 2 is 6

=> (2x-3x)/6 > 1

=> -x/6 > 1

On multiplying with 6 both sides then

=> 6(-x/6) > 1×6

=> -6x/6 > 6

=> -x > 6

=> x < -6

Therefore, x = -7,-8,...∞

The possible values of x are ={-7,-8,..-∞}

Answer:&

The solution set = { -7,-8,...∞}

Answer 2

■Answer:

$=\frac{x}{3} >\frac{x}{2}\;+\;1\\\\=\frac{x}{3}-\frac{x}{2} >1\\\\=\frac{2x-3x}{6}>1\\\\=\frac{-x}{6}>1\\\\= -x>6$

☆Since x negative both side by -1 and change the signs

$=(-1)\;x \; (-x)<(-1)\;x\;(6)\\\\=x<-6$

☆Hence is a real number which is less than -6

$Thus\;solution\;is$ (-∞,-6)

Step-by-step explanation:

☆Hope it's helpful to you☆