Compute the number of ordered pairs of positive integers (x,y) such that 3x + 5y = 500 Hint: you do not want to list them all, but you want to find a creative way to count how many first-quadrant ordered pairs satisfy this equation.
i know it says creative way but...
how do I find them ALLL??
is it like, find the x and y intercepts and then the slope and then just find your way from there??
what happens when x = 0 ?
y = 100
yes
when y is 0, x is ...
so (0,100) is one solution BUT x is not positive here
oh
what is the slope of this line?
um 5y= -3x+500 y= -3/5x + 100
so slope is -3/5x
just -3/5 it's the number out front of the x
oh wait, yeah, oops
slope = -3/5 = rise/run = (change in y)/(change in x) what this means is that when x increases by 5, y decreases by 3 so (0,100) would turn into (5,97) following the rule above notice how 0 turned into 5 (x increased by 5) notice how 100 turned into 97 (y decreased by 3)
mmmhmmm
what about..
the question is: when does y become negative?
when... 33 ordered pairs later..?
but during that, x becomes negative
more like 34 100 - 34*3 = 100 - 102 = -2
oh I see what you mean
oh, yes
so would so since x becomes negative faster than x, do i do something like 100-5x or something?
agh wait, i meant
so 100- 20x5...?
and then taking out the intercepts would mean I have 18 ordered pairs
to generate any ordered pair (x,y) where you want this point to be in Q1, use this rule x = 0 + 5*k y = 100 - 3*k k is an integer k ranges from k = 1 to k = 33 if k < 1 or k > 33, then (x,y) will not be in Q1 (it will be on an axis or in another quadrant)
so this shows that there are 33 ordered pairs (x,y) where x and y are positive integers
ohwait but what about the x being negative?
that happens when k < 1 which I mentioned above
oh, ok
thank you (like usual)
np
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