That's "successive approximation". As a solving method for humans it kinda sucks. For computers, it makes some pretty tidy algorithms.
@ejk Formally, yes, it’s a quadratic. On the other hand, it’s obvious by inspection that the divisor is 9 & the quotient is 2.
@ejk I got it, but the problem I see with it is it leads them to make some assumptions you shouldn't make about the problem, and it doesn't give you those as conditions of the solution to lead you to make them.
Like, I got the answer by assuming that since it's for 3rd grade (which I assume, since my 8yo is in 3rd), they're not using fractional numbers anywhere, so you're limited to whole factors of 18.
@ejk The ones I could think of are 1 & 18, 2 & 9 and 3 & 6. Of those, the only two that are 7 apart are 2 and 9, so that must be it; problem should read:
18 ÷ 9 = 2
And like, the only way I can see to tell the kid to get there is "list all the factor pairs of the dividend, then see if one of the pairs matches up to the problem." I guess all they're wanting to do is get them used to factoring w/ whole numbers? But knowing fractions are to come I feel like this isn't a great way to go about it!
@savedr this is a good explanation, thanks!
@ejk Now that I think about it, it's really a system of equations, if you wanted to graph it out, and they probably intersect on just one point:
18 ÷ x = y and
x - y = 7
So, the first is already pretty simple, y = 18 ÷ x. Let me solve the 2nd for y:
-y + x = 7
-(-y) = -(7 - x)
y = -7 + x, or y = x - 7
The only point that satisfies both equations is (9, 2). So there is just the one solution, including fractional factors!
But of course, they won't see this method till high school Geometry. :D
@ejk Of course, it's really easy to, say, transpose one of the equations so the only solution you get is a fractional one & not the pair I said, like I did when I first typed up that post..!
@ejk ITT: I mansplain solving systems of equations to an electrical engineer with at least one PhD on an instance called "mathstodon"
Loving to explain things is a hell of a drug
@savedr I just love how this question can be made SO COMPLICATED, and I had such trouble thinking of the “third grade” answer
@ejk You could get her to list out all possible divisor/quotient pairs for that dividend and find the pair that works.
@ejk Oh, someone already suggested that.
A Mastodon instance for maths people. The kind of people who make \(\pi z^2 \times a\) jokes.
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