# 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.

2
by Bea6

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by Bea6

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Y-Second Number

Z-Third number

Y=1 + 2z

x=8 + 3y

x + y+ z = 129

Solving the three equations:

X=89

Y=27

Z=13

Therefore the answer is X=89.

The Brainliest Answer!

'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

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Substitute y=27 to equation 1'

z = (y-1)/2

z = (27-1)/2

z = 26/2

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Substitute y=27 to equation 2

x = 3y + 8

x = 3(27) + 8

x = 81 + 8

Therefore the largest number is