Yes, the partition gives the two brothers equal shares.
Step-by-step explanation:
Step 1:
Assume the entire field has an area of B. So one brother takes [tex]\frac{1}{12} B[/tex], [tex]\frac{1}{6}B[/tex], and [tex]\frac{1}{4} B[/tex]
So we need to calculate how much this brother takes in terms of B.
To do this we calculate how much [tex]\frac{1}{12} B + \frac{1}{6} B + \frac{1}{4} B[/tex] is.
Step 2:
To add [tex]\frac{1}{12} B + \frac{1}{6} B + \frac{1}{4} B[/tex],
First take the LCM of the denominators 12, 6, and 4
The LCM is 12, we multiply the denominator to get the LCM value, this same value is multiplied to the numerator too.
[tex]\frac{1}{12} B + \frac{1}{6} B + \frac{1}{4} B = \frac{1(1)}{12(1)} B + \frac{1(2)}{6(2)} B + \frac{1(3)}{4(3)} B \\\\\frac{1}{12} B + \frac{1}{6} B + \frac{1}{4} B= \frac{1B +2B+3B}{12} = \frac{B}{2}[/tex]
Step 3:
One brother gets [tex]\frac{1}{2} B[/tex], so we need to calculate how much the other brother gets.
The other brother's share = [tex]B - \frac{1}{2} B = \frac{1}{2}B[/tex]
So both the brothers get equal shares
