If you have x $5 notes and y $10 notes, what is their total value?I tried to setting up simultaneous equations
x for number of $5 notes
y for number of $10 notes
is there a third variable for the total?
how do I find total possible combinations?
is using a graph helpful at all?
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5x +10y = ? it will still be 5x +10y right?If you have x $5 notes and y $10 notes, what is their total value?
The graph may help a little in imagining the problem, but since x and y must be positive integers, it will not solve the problem by itself.
Do you know anything about Diophantine equations? You're look for all the integer solutions within the triangle.
No, I just asked because knowing your level of knowledge can be very important, and because it's a relevant term; I don't have any advanced methods in mind.5x +10y = ? it will still be 5x +10y right?
I have no clue about diophantine equations , I am still 16 years old, not a university student... is this supposed to be advance?
Multiples of 5 is the main "trick" I had in mind to save work. That makes it a little better than mere brute force. (I assume you mean that you recognized that at least one of the variables has to be a multiple of 5.)I can solve simultaneous equations using elimination & substitution method, but i can't seem to do both here, can you give me more hint? or simpler explanation? should i just try brute force in trial & error?
I did try many values of x and y in multiples of 5, but still unable to find the values that will fit xy=5x+10y, any hint on the process?
do you mean make x or y as the subject of the equation?Can you solve for one variable in terms of the other in xy=5x+10y as requested by Dr Peterson?
Trial and error is a legitimate mathematical procedure - unless explicitly forbidden by the instructor.no mathematical algebra to solve the answer?
Share the combinations you have tried.can somebody really help me in getting the combination, i am still stuck honestly despite trying several combinations
yes i just tried y = (5x) / (x-10)Share the combinations you have tried.
Did you write an expression for 'y' - in terms of 'x'?
oh wait i put them in an organised table
first column is x and second column is y
then i have shorter possible list of multiples of 5
i check that x=20 and y=10 is the possible answer
they match, is it correct?
Yes, that is one of several answers; you have to find all of them.oh wait i put them in an organised table
first column is x and second column is y
then i have shorter possible list of multiples of 5
i check that x=20 and y=10 is the possible answer
they match, is it correct?
oh bummer, I just realised that I got one answer only out of several possible answers.... so to find the rest, i really have to "observe and match" from a list or a table? there is really no algebraic or other mathematical methods available?Yes, that is one of several answers; you have to find all of them.
I did what you did, using the formula to make it easier to create the table, using only multiples of 5 for x.
But I also had to solve for x in terms of y, and repeat the table in the other direction, because it isn't always x that is a multiple of 5. In all, I made 15 rows to check (though some didn't need to be actually calculated.
Yes. There is nothing wrong with trying all cases, as long as you have used reasoning to reduce their number. This is not at all uncommon even after you learn about Diophantine equations and methods for solving them!oh bummer, I just realised that I got one answer only out of several possible answers.... so to find the rest, i really have to "observe and match" from a list or a table? there is really no algebraic or other mathematical methods available?