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How do you solve #38+5x >7 (x+4)#?

Answer 1

Nathan Axel

#x<5#

#38+5x>7x+28#

#5x-7x> -38+28#

#-2x> -10#

#2x<10#

#x<5#

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Answer 2

Eva Abramson

#x<5#

First, let's switch the sides:

#7(x+4)<5x+38#

Next, we can distribute the #7# on the left to get

#7x+28<5x+38#

Next, subtract #5x# from both sides to get

#2x+28<38#

We can subtract #28# from both sides to get

#2x<10#

Lastly, we divide both sides by #2# to get

#x<5#

Hope this helps!

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Answer 3

Hazel Anglin

To solve the inequality 38 + 5x > 7(x + 4):

Distribute 7 across (x + 4) on the right side: 38 + 5x > 7x + 28
Move variables to one side and constants to the other: 38 - 28 > 7x - 5x
Simplify both sides: 10 > 2x
Divide both sides by 2 to solve for x: 10/2 > 2x/2
Simplify: 5 > x

So, the solution is x < 5.

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Answer 4

Jameson Allen

To solve the inequality ( 38 + 5x > 7(x + 4) ):

Distribute 7 on the right side: ( 38 + 5x > 7x + 28 ).
Subtract ( 7x ) from both sides: ( 38 - 2x > 28 ).
Subtract 38 from both sides: ( -2x > -10 ).
Divide both sides by -2 (note that dividing by a negative number reverses the inequality): ( x < 5 ).

So, the solution to the inequality is ( x < 5 ).

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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 #38+5x &gt;7 (x+4)#?

How do you solve #38+5x >7 (x+4)#?