(30 * 5) - (10n) >= 100
30 * 5 - 10n >= 100
150 - 10n >= 100
-150 + 150 -10n >= 100 - 150
-10n >= -50
-10n /-10 >= -50/-10
n <= 5
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(30X5) - (10n) >= 100
Problem. Isolate n to determine max num of refuels.
150 - 10n >=100
-10n >= -150 + 100
-10n/-10 >= -50/-10
n = 5
A max number of 5 refuels are allowed.
Your answer is correct! But your notation is slightly off. n <= 5 but indeed the maximum numbers of refuels is correct!
Kind regards,
Lisa
Hey, you’re right! Thanks for the correction!
Jeff
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(30 x 5 ) - (10x) >= 100
150 - 10 <= 10x
50 <= 10x
50 / 10 >= n
= 5 >= n
(30 * 5) - (10n) >= 100
150 - 10n >= 100
n <= 5
I’m frustrated. Got the answer was 5 right away but that wasn’t the point of the question.
Made a note to study algebra some more later.
(30*5) - (10n) >= 100
(150) - (10n) >= 100
Subtract 150 from both sides
150(-150) - (10n) >= 100(-150)
-10n >= -50
Divide both sides by -10 to isolate the n. Flip the sides because negative division.
-10(/-10) >= -50(/-10)
n <= 5
Another fun one. I appreciated the hint starting point 
n <= 5
Meaning the player can refuel up to 5 times and still be able to afford an upgrade.
150 - 10n >= 100
-10n >= -50
-10n/-10 <= -50/-10
n <= 5
(30*5) - (10n) >= 100
150 - 100 >= 10n
50 / 10 >= n
n <= 5
(30 * 5) - 10n >= 100
Summary
150 - 10n >= 100 | -150
-10n >= -50 | / (-10)
n <= 5
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