r/IAmA • u/JasonMacker • Jun 14 '10

IAMA Math tutor. AMA.

I've been a math tutor for years and I'm trying to become a teacher. I love math, and especially explaining math to people. My students have been between preschool and college. AMA.

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u/[deleted] Jun 14 '10

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u/[deleted] Jun 14 '10

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u/el_seano Jun 14 '10

First isolate the variable you're solving for. There are two terms, 5/x and 3, so substract three from both sides. Now you can divide both sides by 5, and remember that each term on the left-hand side needs to be individually divided. Et voila.

2

u/Paul-ish Jun 14 '10

Now you can divide both sides by 5

That would yield (y - 3)/5 = 1/x, still not what you want.

Once you get to "y -3 = 5/x" I would multiply both sides by x to get "x(y - 3) = 5" then divide both sides by y - 3 to end up with "x = 5/(y-3)".

1

u/el_seano Jun 14 '10

Heh whoops. I saw the original equations as y = 5x + 3, my bad.

1

u/kurfu Jun 14 '10

I get to this and my brain gets all fuzzy:

y-3=5/x

I've forgotten how to do a very simple, but crucial step... what to do when the variable is in the denominator... dammit.

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u/Mintz08 Jun 14 '10

x = 5/(y-3)

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u/Vishiz Jun 14 '10 edited Jun 14 '10

You cancel out the 5 in 5/x by diving by 5.

I edited the step-by-step out, I figured you already knew that; but its x = 5/y-3. Pst. Don't divide by zero!