Question

for any x, 3 x + 1 is greater than 4​

Answer 1

question is

x(x-1)(x+2)>0

Firt ignore the '>' sign:

so: x(x-1)(x+2)=0

Since 3 numbers are multiplied to give 0, then each may potentially be zero, so:

x = 0

x - 1 = 0, which means x = 1

x + 2 = 0, which means x = -2

So, you have 3 numbers -2, 0, 1

Now, draw a straight line and mark these 3 points on the line. You can see now that it these numbers have divided the line into 4 different regions. Region 1 is (-inf, -2) -- that is all numbers to the left of -2.

Region 2 is (-2,0) --- that is all numbers between -2 and 0

Region 3 is (0,1) --- that is all numbers between 0,1

Region 4 is (1,inf)--- that is all numbers to the right of 1

Your strategy now, is to test a number inside of each of these 4 regions. Basically you pick a number inside, but not the end values. You substitute the number into the original inequality for x, and see which ones are true.

Hence:

Region 1: (-inf, -2),

I can choose -5 since its in the region to the left of -2.

I get:

(-5)(-5-1)(-5+2) >? 0 I used '?' because I am not sure if it is true

so: (-5)(-6)(-3) >? 0

Finally -90 >? 0 is false. SO Mark this region with an X on your line

Answer 2

Given,

3x + 1 > 4

-» 3x > 4 - 1

-» 3x > 3

-» x > 1

Therefore, x is greater than 1 .