How do you solve #3(x + 2) > x#?

Answer 1

x>-3.

#3(x+2) > x# => #3x+6 > x# => #3x+6-x > x-x# ; (subtract x from both sides) =>#2x+6 > 0# =>#2x+6-6 > 0-6#; (subtract 6 from both sides) => 2x > -6 now finally divide by 2 on both sides to get, => #x > -3#

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Answer 2

Julia Adams

#x> -3#

Expand:

#3(x+2)>x -> 3x+6>x#

Get #x# on one side, then minus #6#

#3x+6>x -> 2x> -6#

Divide by #2#:

#x> -3#

Remember what we do on one side, we do to the other. Use this when doing all the steps above

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Answer 3

Ethan Adkins

To solve ( 3(x + 2) > x ), distribute 3 across ( x + 2 ) and then solve for ( x ):

[ 3x + 6 > x ]

Subtract ( x ) from both sides:

[ 2x + 6 > 0 ]

Subtract 6 from both sides:

[ 2x > -6 ]

Divide both sides by 2:

[ x > -3 ]

So the solution is ( x > -3 ).

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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 #3(x + 2) &gt; x#?

How do you solve #3(x + 2) > x#?