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THE 2ND NO. IS 1 MORE THAN TWICE OF THE THIRD NO. AND THE FIRST NO. IS EIGHT MORE THAN 3X OF THE SECOND NO.IF THERE SUM IS 129, WHAT IS THE LARGEST NO.

Let 'x' be the 1st number 'y' be the 2nd number 'z' be the 3rd number -------------------------------- -The 2nd number is 1 more than twice the third number y = 2z + 1 -----equation 1 ------------------------------- -The 1st number is 8 more than thrice of the 2nd x = 3y + 8 -----equation 2 ------------------------------ -Their sum is 129 x + y + z = 129 ---equation 3 ----------------------------- Expressing everything in terms of y, you'll have: From equation 1 y = 2z + 1 2z = y - 1 z = (y-1)/2 ------equation 1' ---------------------------- Substitute equations 1' and 2 to equation 3 x + y + z =129 (3y+8) + y + (y-1)/2 = 129 Multiply the whole equation with 2 6y + 16 + 2y + y - 1 = 258 6y + 2y + y = 258 + 1 - 16 9y = 243 y = 27 -------------------------- Substitute y=27 to equation 1' z = (y-1)/2 z = (27-1)/2 z = 26/2 z = 13 -------------------------- Substitute y=27 to equation 2 x = 3y + 8 x = 3(27) + 8 x = 81 + 8 x = 89 -------------------------- Therefore the largest number is 89.