Question 999176
ax + by = c


the x-intercept is the value of x when y = 0.
the y-intercept is the value of y when x = 0.


when y = 0, ax = c which results in x = c/a


when x = 0, by = c which results in y = c/b


in order for c/a to be an integer, c has to be a multiple of a.


in order for c/b to be a fraction, c must not be a multiple of b.


if c is a multiple of a, then c can be made equal to a*d, where d is an integer.


example:


55 / 5 is an integer because 55 is a multiple of 5.
55 = 5 * 11, so 55/5 = 11


56 / 5 is not an integer because 56 is not a multiple of 5.


-15 / 5 is an integer because 15 is a multiple of 5.
15 = 5 * 3, so 15/5 = 3


-16 / 5 is not an integer because 16 is not a multiple of 5.


so, if you want ax + by = c to have the x-intercept an integer and the y-intercept a fraction, then c must be a multiple of a but not a multiple of b.


an example would be:


5x + 6y = 15


the x intercept is the value of x when y = 0.
when y = 0, the equation becomes 5x = 15.
solve for x to get x = 15/5 = 3.
3 is an integer.


the y intercept is the value of y when x = 0.
when x = 0, the equation becomes 6y = 15.
solve for y to get y = 15/6.
15/6 is not an integer.


that's my take.


an integer can also be described as a fraction where the denominator is equal to 1.


why is 55/5 an integer?


multiply numerator and denominator of 55/5 by 1/5 and you get (55*1/3) / (5*1/5) which results in 11/1 which is the same as 11.


55/5 is therefore an integer because it can be simplified to a ratio where the denominator is equal to 1.


56/5 cannot be simplified to a ratio where the numerator is an integer and the denominator is equal to 1.


this is because 56 is not a multiple of 5.


you can find a number such that the denominator is 1, but the number will not be an integer.


56/5 = 11.2/1 = 11.2 which is a fraction in decimal form.