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How do you solve #-2abs(x-3)+6 < -4#?

Answer 1

Hazel Anglin

There are two possibilities: #x-3# (without the absolutes) is positive or negative (the 'switch point is #x=3#)

If it is positive, we can leave out the absolute brackets, is it is negative, we 'turn around' to #-(x-3)=3-x#

(1) #x-3>=0->x>=3# #-2(x-3)+6<-4->-2x+6+6<-4-># #-2x<-16->x>8#

(2)#x-3<0->x<3# #-2(3-x)+6<-4-> -6+2x+6<-4-># #2x<-4->x<-2#

Conclusion: #x<-2orx>8#

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Answer 2

Genesis Allen

To solve the inequality, follow these steps:

Subtract 6 from both sides: -2|𝑥−3| < -4 - 6
Simplify: -2|𝑥−3| < -10
Divide both sides by -2. Since we are dividing by a negative number, flip the inequality sign: |𝑥−3| > 5
Split into two cases: 𝑥−3 > 5 or 𝑥−3 < -5
Solve each case separately: For 𝑥−3 > 5: 𝑥 > 5 + 3 𝑥 > 8

For 𝑥−3 < -5: 𝑥 < -5 + 3 𝑥 < -2

Combine the solutions: 𝑥 > 8 or 𝑥 < -2

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you solve #-2abs(x-3)+6 &lt; -4#?

How do you solve #-2abs(x-3)+6 < -4#?