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OpenStudy (anonymous):
i still don't understand my own questin
OpenStudy (anonymous):
um, i am sure amistre can help further
OpenStudy (anonymous):
he is good with explaining
OpenStudy (anonymous):
hey @mads4566 since you are only given one equation you have to solve this by trial and error
OpenStudy (amistre64):
ok, my first question is; what was the original question?
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OpenStudy (anonymous):
3x+2y=25 there are infinite many solutions how many solutions are you looking for?
OpenStudy (anonymous):
look at that first question as two different questions:
X 2 5 2
--- = --- AND --- = ---
5 3 Y 3
OpenStudy (anonymous):
x/5=2/3=5/y is the originl
OpenStudy (anonymous):
i have to find x and y
OpenStudy (amistre64):
so its a group of proportions is what it looks like to me
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OpenStudy (amistre64):
cross multiplying will seem like a good bet
OpenStudy (anonymous):
OO so x=10/3 and y = 15/2
OpenStudy (anonymous):
use that equation maker thingy :)
OpenStudy (anonymous):
how could x possibly be 10/3?
OpenStudy (anonymous):
yeah @amistre64
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OpenStudy (amistre64):
x/5=2/3 ; multiply by 3*5 to clear the fractions
3x = 10; divide off the 3
x = 3/10
2/3=5/y ; multiply thru by 3*y to clear the fractions
2y = 15; divide off teh 2
y = 15/2
OpenStudy (anonymous):
so i need to sollve for one at a time instead or extending cross multiplication ... like x/5=2/3 and solve for x
OpenStudy (anonymous):
your x is wrong bro
OpenStudy (amistre64):
its simpler to do them one at a time
OpenStudy (anonymous):
Oohhh
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OpenStudy (amistre64):
yeah, typoed it :) x = 10/3
OpenStudy (anonymous):
Thank you so much!! :)
OpenStudy (amistre64):
youre welcome, the idea is to make them all equal, and since you have a solid value to work with; equate the unknowns to that one