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How do you solve #6( 2x + 5) < - 78#?

Answer 1

Eva Adamson

#x< -9#

#"distribute the parenthesis"#

#rArr12x+30< -78#

#"subtract 30 from both sides"#

#12xcancel(+30)cancel(-30)< -78-30#

#rArr12x< -108#

#"divide both sides by 12"#

#(cancel(12) x)/cancel(12)<(-108)/12#

#rArrx< -9" is the solution"#

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Answer 2

Ethan Adkins

To solve the inequality (6(2x + 5) < -78), follow these steps:

Distribute the 6 on the left side: (12x + 30 < -78)
Move the constant term to the right side: (12x < -78 - 30) (12x < -108)
Divide both sides by 12 (since 12 is positive, inequality direction remains unchanged): (x < -108/12) (x < -9)

The solution to the inequality is (x < -9).

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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 #6( 2x + 5) &lt; - 78#?

How do you solve #6( 2x + 5) < - 78#?