Hi, can you explain how to get the equation for test 6, section 4, number 31, or fix the page error?

Yes, and sorry about that page error. I am not sure why a very small number of old links stopped working!

The way to get this is to say that the original number of friends intending to go on the trip is n. So the original amount each friend planned to pay was \dfrac{800}{n}.

When two friends decide not to go, the price of the trip doesn’t change but now there are two fewer friends, so now each friend is going to pay \dfrac{800}{n-2}.

The question tells us that in this second scenario, the friends will each pay $20 more, so we can write an equation and solve for n.


To solve, multiply through by both denominators.

\left(\dfrac{800}{n}+20\right)(n)(n-2)=\left(\dfrac{800}{n-2}\right)(n)(n-2)\\\\800(n-2)+20n(n-2)=800n\\\\800n-1600+20n^2-40n=800n\\\\20n^2-40n-1600=0\\\\n^2-2n-80=0\\\\(n-10)(n+8)=0\\\\n=10 \text{ or }n=-8

You can’t have a negative number of friends, so there were originally 10 friends (each paying $80) and then when 2 friends bailed, there were 8 friends paying $100 each.

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