Well theoretically speaking you could but that would take a long time and 15 and 30 have a common factor of 5 and 15. So I think for this one you can't since there are two common factors.
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Answer:
The integer 27 is an odd number. The integer 27 is a Composite number. 13 is less than 27, so 27 is a deficient number.
(x + a)/(ax) = b/x
a= 2
b = 5
(x + 2) /(2x) = 5/x
cross multiply
x * (x+2) = (2x)*5
x^2 + 2x = 10x
x^2 - 8x = 0
x ( x -8) = 0
We get two solutions
x = 0, 8
the solution 0 is extraneous
the solution 8 is valid.
Answer:
0.5×2/2=1/2 is the required fraction.
