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06-17-2005, 08:21 AM
I need help solving this

x/2 = 3y/5

tkhunny
06-17-2005, 10:35 AM
We seem to be having trouble with language, today. What does "solving this" mean? Solve for 'x'? Solve for 'y'? Find simultaneous solutions for particular 'x' and 'y'? An example of what you are doing would be very beneficial.

x/2 = 3*y/5

x/2 must be an integer, so 'x' must be divisible by 2.

3 and 5 are prime, so let's ignore the 3 for a moment.

y/5 must be an integer, so 'y' must be divisible by 5

Solve for x

x = 6*y/5

Pick a 'y' that is divisible by 5.

x = 6*5/5 = 6

x = 6*0/5 = 0

OK, that was fun. Now, what was the problem we were supposed to work on?

06-17-2005, 11:26 AM
I am working equations of lines in slope intercept using a graph. I am to write each equation in standard form using only integers. Does this explain it?

Problem is x/2 = 3y/5

tkhunny
06-17-2005, 12:16 PM
I am to write each equation in standard form using only integers.
THERE is the missing clue.

"Standard Form" is usually something like:

Ax + By + C = 0, where A, B, and C are integers. Or, maybe it is OK to have Ax - By, but that is of little consequence. (Notice how this does not allow such things as the square root of two.)

You just need the least common denominator and a little algebra.

2 and 5 -- The LCD is 10

x/2 = 3y/5
10*x/2 = 10*3y/5
(10/2)*x = (10/5)*3y
(5)*x = (2)*3y
5*x = 2*3y
5*x = 6y
5*x - 6y = 0