How do you solve the compound inequalities #3x≥-12# and #8x≤16#?

Answer 1
Taking the two inequalities separately #3x>= -12# #rarr x>= -4#
#8x<=16# #rarr x<=2#
Combining: #3x>=-12" and " 8x<=16# #rarr -4 <= x <= 2#
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Answer 2
(1) # 3x >= - 12 -> x >= -4#
(2) # 8x <= 16 -> x <= 2#
Compound solution set : #-4 <= x <= 2#

The solution set is the close interval: [-4, 2]. The 2 end points are included in the solution set.

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Answer 3

To solve the compound inequalities (3x \geq -12) and (8x \leq 16), you solve each inequality separately and then find the intersection of the solution sets. So, for (3x \geq -12), you divide both sides by (3) to isolate (x), yielding (x \geq -4). And for (8x \leq 16), you divide both sides by (8), giving (x \leq 2). The intersection of these solution sets is the range where both inequalities hold true, which is (-4 \leq x \leq 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.

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.

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.

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.

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