If |3x + 7| ≥ 2x + 12, then which of the following is true?

|3x+7| \(\geq{2x+12}\)

The critical points of the above equation are -6 and \(\frac{-7}{3}\)
To find the critical Points equate the either side of the equations to 0
i.e. 3x + 7 = 0 ==> x =\(\frac{-7}{3}\)
and 2x + 12 = 0 ==> x = -6

So 3 ranges to consider :
Range 1 : x<-6
Take any value in this area say for Ex : -7

Look for the Expr. inside mod 3x +7 will become 3(-7)+7
So this expression will evaluate to a -ve value

Therefore this equation can be written as :

(-1)(3x+7) \(\geq{2x+12}\)
-3x-7 \(\geq{2x+12}\)
-5x \(\geq{19}\)
x \(\leq{-19/5}\)

Now this range is x<-6 , so based on the solution all values of x in this range should satisfy the equation

Range 2 : -6<x<\(\frac{-7}{3}\)
Take any value in this area say for Ex : -5

Look for the Expr. inside mod 3x +7 will become 3(-5)+7
So this expression will evaluate to a -ve value

Therefore this equation can be written as :

(-1)(3x+7) \(\geq{2x+12}\)
-3x-7 \(\geq{2x+12}\)
-5x \(\geq{19}\)
x \(\leq{-19/5}\)

Now this range is -6<x<\(\frac{-7}{3}\) , so only values of x \(\leq{-19/5}\) will satisfy

Range 3 : x>\(\frac{-7}{3}\)
Take any value in this area say for Ex : -1

Look for the Expr. inside mod 3x +7 will become 3(-1)+7
So this expression will evaluate to a +ve value

Therefore this equation can be written as :

(3x+7) \(\geq{2x+12}\)
3x+7 \(\geq{2x+12}\)
3x-2x \(\geq{5}\)
x \(\geq{5}\)

Now this range is x>\(\frac{-7}{3}\), so only values of x\(\geq{5}\) will satisfy

So Combining ,x \(\leq{-19/5}\) OR x \(\geq{5}\) will hold true

Hence Answer : D

Can you please correct me?

case 1:
|3x+7|>= 2x+12
x>=5

case 2:
|3x+7|>= -(2x+12)
3x+7>=-2x-12
3x+2x>=-12-7
5x>=-19
x>=-19/5

so,
x>=5 and x>= -19/5

Where I am wrong?

You are wrong in case II.
You have to take negative of the MOD and not the other side
So -(3x+7)>=2x+12
If I have to carry negative sign on other side, I will have to change the INEQUALITY sign.
So (3x+7)<=-(2x+12)

Thank you so much, I could not recall the changing of direction of sign.
One quick question, does it happen in every case?
like if |x|<=7
then x<=7 and x>=-7
right?

Yes
|x|<7....-7<x<7
|x|>7...x>7 and x<-7

Hi,
In your first method, Is it possible to square both sides without knowing the sign of (2x+12)?

By squaring on both sides I am getting the answers x≥5 and x≥−19/5

Could someone please explain where I am going wrong in the 2nd part of the answer?

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