Given:
LCM(a,b) = 90 and GCD(a,b) = 3.
b is three more than a.
To find:
The values of a and b.
Solution:
We have,
LCM(a,b) = 90
GCD(a,b) = HCF(a,b) = 3
According to the question,
[tex]b=a+3[/tex]
If a and b are two positive integers, then
[tex]a\times b=HCF(a,b)\times LCM(a,b)[/tex]
[tex]a\times (a+3)=3\times 90[/tex]
[tex]a^2+3a=270[/tex]
[tex]a^2+3a-270=0[/tex]
Splitting the middle terms, we get
[tex]a^2+18a-15a-270=0[/tex]
[tex]a(a+18)-15(a+18)=0[/tex]
[tex](a+18)(a-15)=0[/tex]
Using zero product property, we get
[tex]a+18=0[/tex] and [tex]a-15=0[/tex]
[tex]a=-18[/tex] and [tex]a=15[/tex]
a is a positive integer so it cannot be negative. So, a=15.
Now,
[tex]b=a+3[/tex]
[tex]b=15+3[/tex]
[tex]b=18[/tex]
Therefore, the value of a is 15 and the value of b is 18.
