The sum of the ages of Berma, her mother Rinna and her father Erwin is 80. Two years from now, Rinna’s age will be 13 less than the sum of Erwin’s age and twice Berma’s age. Three years ago, 15 times Berma’s age was 5 less than the age of Rinna. How old are they now?

Answer: Berma is 5 years old Rinna is 38 years oldErwin is 37 years old Step-by-step explanation:Let x represent Berma's ageLet y represent Rinna's ageLet z represent Erwin's ageSince the sum of their ages is 80,x + y + z = 80 - - - - - - -1Two years from now, Rinna’s age will be 13 less than the sum of Erwin’s age and twice Berma’s age. This means thaty +2 = [ (z+2) + 2(x+2) ] - 13y +2 = z + 2 + 2x + 4 - 132x - y + z = 13 + 2 - 4 -22x - y + z = 9 - - - - - - -2Three years ago, 15 times Berma’s age was 5 less than the age of Rinna. It means that15(x - 3) = (y - 3) - 515x - 45 = y - 3 - 515x - y = - 8 + 4515x - y = 37 - - - - - - - -3From equation 3, y = 15x - 37Substituting y = 15x - 37 into equation 1 and equation 2, it becomesx + 15x - 37 + z = 8016x + z = 80 + 37 = 117 - - - - - - 42x - 15x + 37 + z = 9-13x + 2 = -28 - - - - - - - - -5subtracting equation 5 from equation 4,29x = 145x = 145/29 = 5y = 15x - 37 y = 15×5 -37 y = 38Substituting x= 5 and y = 38 into equation 1, it becomes5 + 38 + z = 80z = 80 - 43z = 37