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Solve equation of two variables, only one equation

I am given that

m×8+n×7=0 m \times 8 + n \times 7 = 0

There are 2 variables here. In my specific problem I only need integer solutions. What is the process of finding solutions in this case. Do you just guess? ie m = 7 and n = -8 fit??? Or is there a method for doing this?
Do trial and error maybe?
rearrange to get m= or
input integers in n, and see if it is an integer.

however this is a very simple one, and to me the solution is obvious just by doing 8m=7n

BUT there are an infinite amount of solutions, just multiply and n by an integer
Reply 2
Original post by acomber
I am given that

m×8+n×7=0 m \times 8 + n \times 7 = 0

There are 2 variables here. In my specific problem I only need integer solutions. What is the process of finding solutions in this case. Do you just guess? ie m = 7 and n = -8 fit??? Or is there a method for doing this?


You are correct, there are an infinite number of solutions. I find that the best way to understand it is to rearrange the problem so..
[br]8m+7n=0[br]8m=7n[br]mn=78 [br]8m + 7n = 0[br]\Rightarrow 8m = -7n[br]\Rightarrow \frac{m}{n} = \frac{-7}{8}

Now, just pick any m and n such that we get that fraction. i.e choosing m= -7 and n = 8 is a trivial solution, but note that for any non zero constant c, say, we can have m= -7c and n=8c and they will also be solutions since the fraction still holds true (as m/n is still the required fraction since c is non zero)
For example, take c=2 , then we have m=-14 and n=16. This is also a solution to the equation.
Does this help?
(first time using LaTeX btw..)
Original post by acomber
I am given that

m×8+n×7=0 m \times 8 + n \times 7 = 0

There are 2 variables here. In my specific problem I only need integer solutions. What is the process of finding solutions in this case. Do you just guess? ie m = 7 and n = -8 fit??? Or is there a method for doing this?
Given

Am + Bn = 0

Then if A, B have no common factors the solution set is {m=kB,n=kA,kZ}\{m = kB, n = -kA, k \in \mathbb{Z} \}.

If A, B have common factors then the solution set is {m=kB/C,n=kA/C,kZ}\{m = kB/C, n = -kA/C, k \in \mathbb{Z} \}, where C = hcf(A, B).

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