(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.

(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

150 - 10n >= 100 | -150

-10n >= -50 | / (-10)

n <= 5

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(150) - (10n) >= 100

-(10n) >= 150 - 100

10n >= 50

n <= 50/10

n <= 5

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(30*5) - (10n) >= 100

<=> 150 - 10n >= 100

<=> -10n >= -50

<=> n <= 5

1 Like