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How do you solve #-3x + 2 >11# or #5(x - 2) <0#?

Answer 1

Ethan Adkins

Solution : # x < -3 # , in interval notation : # (-oo, -3)#

1) # -3x +2 > 11 or -3x > 9 or -x > 3 or x < -3 #

2) # 5 (x-2) < 0 or x-2 < 0 or x < 2#

# x < -3 # satisfies both inequaity.

Solution : # x < -3 # , in interval notation : # (-oo, -3)# [Ans]

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Answer 2

Avery Alexander

To solve the inequality -3x + 2 > 11 or 5(x - 2) < 0:

For -3x + 2 > 11: -3x + 2 > 11 -3x > 11 - 2 -3x > 9 x < 9/-3 x < -3

For 5(x - 2) < 0: 5(x - 2) < 0 x - 2 < 0 x < 2

So, the solution to the inequality -3x + 2 > 11 or 5(x - 2) < 0 is x < -3 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 #-3x + 2 &gt;11# or #5(x - 2) &lt;0#?

How do you solve #-3x + 2 >11# or #5(x - 2) <0#?