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